John is a Chief Executive Officer at a privately owned medium size company. The owner of the company has decided to make his son Scott a manager in the company. John fears that the owner will ultimately give CEO position to Scott if he does well on his new manager position, so he decided to make Scott’s life as hard as possible by carefully selecting the team he is going to manage in the company.

John knows which pairs of his people work poorly in the same team. John introduced a hardness factor of a team — it is a number of pairs of people from this team who work poorly in the same team divided by the total number of people in the team. The larger is the hardness factor, the harder is this team to manage. John wants to find a group of people in the company that are hardest to manage and make it Scott’s team. Please, help him.

In the example on the picture the hardest team consists of people 1, 2, 4, and 5. Among 4 of them 5 pairs work poorly in the same team, thus hardness factor is equal to ⁵⁄₄. If we add person number 3 to the team then hardness factor decreases to ⁶⁄₅.

The first line of the input file contains two integer numbers n and m (1 ≤ n ≤ 100, 0 ≤ m ≤ 1000). Here n is a total number of people in the company (people are numbered from 1 to n), and m is the number of pairs of people who work poorly in the same team. Next m lines describe those pairs with two integer numbers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) on a line. The order of people in a pair is arbitrary and no pair is listed twice.

Write to the output file an integer number k (1 ≤ k ≤ n) — the number of people in the hardest team, followed by k lines listing people from this team in ascending order. If there are multiple teams with the same hardness factor then write any one.

Note, that in the last example any team has hardness factor of zero, and any non-empty list of people is a valid answer.

#include <stdio.h>#include <iostream>#include <algorithm>#include <cstdlib>#include <sstream>#include <numeric>#include <cstring>#include <bitset>#include <string>#include <deque>#include <stack>#include <cmath>#include <queue>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)#define CLR(arr,val) memset(arr,val,sizeof(arr))#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);typedef pair<int, int> pii;typedef long long LL;const double PI = acos(-1.0);const int N = 110;const int M = 1010;const double eps = 1e-8;struct edge{    int to, nxt;    double cap;    edge() {}    edge(int _to, int _nxt, double _cap): to(_to), nxt(_nxt), cap(_cap) {}};struct info{    int u, v;};info e[M];edge E[(N + 3 * M) << 1];int head[N + M], tot;int d[N + M];void init(){    CLR(head, -1);    tot = 0;}inline void add(int s, int t, double cap){    E[tot] = edge(t, head[s], cap);    head[s] = tot++;    E[tot] = edge(s, head[t], 0);    head[t] = tot++;}int bfs(int s, int t){    queue<int>Q;    CLR(d, -1);    d[s] = 0;    Q.push(s);    while (!Q.empty())    {        int u = Q.front();        Q.pop();        for (int i = head[u]; ~i; i = E[i].nxt)        {            int v = E[i].to;            if (d[v] == -1 && E[i].cap > 0)            {                d[v] = d[u] + 1;                if (v == t)                    return 1;                Q.push(v);            }        }    }    return ~d[t];}double dfs(int s, int t, double f){    if (s == t || f == 0)        return f;    double ret = 0;    for (int i = head[s]; ~i; i = E[i].nxt)    {        int v = E[i].to;        if (d[v] == d[s] + 1 && E[i].cap > 0)        {            double df = dfs(v, t, min(f, E[i].cap));            if (df > 0)            {                E[i].cap -= df;                E[i ^ 1].cap += df;                ret += df;                f -= df;                if (f == 0)                    break;            }        }    }    if (ret == 0)        d[s] = -1;    return ret;}double dinic(int s, int t){    double ret = 0;    while (bfs(s, t))        ret += dfs(s, t, 1e9);    return ret;}double Maxweight(int n, int m, double g){    init();    int S = 0, T = n + m + 1;    int i;    for (i = 1; i <= n; ++i) //负权点连到汇点T        add(i, T, g);    for (i = 1; i <= m; ++i) //m个正权点从S连出    {        add(S, n + i, 1.0);        add(n + i, e[i].u, 1e9); //原有边保留        add(n + i, e[i].v, 1e9);    }    return m * 1.0 - dinic(S, T); //正权点-最大流得到最大权}int main(void){    int n, m, i;    while (~scanf("%d%d", &n, &m))    {        for (i = 1; i <= m; ++i)            scanf("%d%d", &e[i].u, &e[i].v);        if (!m)        {            puts("1");            puts("1");        }        else        {            double L = 1.0 / n, R = m;            double ans = 1;            double dx = 1.0 / n / n;            while (fabs(R - L) >= dx)            {                double mid = (L + R) / 2.0;                if (fabs(Maxweight(n, m, mid)) >= eps)                {                    L = mid;                    ans = mid;                }                else                    R = mid;            }            Maxweight(n, m, ans);            vector<int>pos;            for (i = 1; i <= n; ++i)                if (~d[i])                    pos.push_back(i);            int sz = pos.size();            sort(pos.begin(), pos.end());            printf("%d\n", sz);            for (i = 0; i < sz; ++i)                printf("%d\n", pos[i]);        }    }    return 0;}

#include <stdio.h>#include <iostream>#include <algorithm>#include <cstdlib>#include <sstream>#include <numeric>#include <cstring>#include <bitset>#include <string>#include <deque>#include <stack>#include <cmath>#include <queue>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)#define CLR(arr,val) memset(arr,val,sizeof(arr))#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);typedef pair<int, int> pii;typedef long long LL;const double PI = acos(-1.0);const int N = 110;const int M = 1010;const double eps = 1e-8;struct edge{    int to, nxt;    double cap;    edge() {}    edge(int _to, int _nxt, double _cap): to(_to), nxt(_nxt), cap(_cap) {}};struct info{    int u, v;};info e[M];edge E[(N + M) << 2];int head[N], tot;int d[N], deg[N];void init(){    CLR(head, -1);    tot = 0;}inline void add(int s, int t, double cap){    E[tot] = edge(t, head[s], cap);    head[s] = tot++;    E[tot] = edge(s, head[t], 0);    head[t] = tot++;}int bfs(int s, int t){    queue<int>Q;    CLR(d, -1);    d[s] = 0;    Q.push(s);    while (!Q.empty())    {        int u = Q.front();        Q.pop();        for (int i = head[u]; ~i; i = E[i].nxt)        {            int v = E[i].to;            if (d[v] == -1 && E[i].cap > 0)            {                d[v] = d[u] + 1;                if (v == t)                    return 1;                Q.push(v);            }        }    }    return ~d[t];}double dfs(int s, int t, double f){    if (s == t || f == 0)        return f;    double ret = 0;    for (int i = head[s]; ~i; i = E[i].nxt)    {        int v = E[i].to;        if (d[v] == d[s] + 1 && E[i].cap > 0)        {            double df = dfs(v, t, min(f, E[i].cap));            if (df > 0)            {                E[i].cap -= df;                E[i ^ 1].cap += df;                ret += df;                f -= df;                if (f == 0)                    break;            }        }    }    if (ret == 0)        d[s] = -1;    return ret;}double dinic(int s, int t){    double ret = 0;    while (bfs(s, t))        ret += dfs(s, t, 1e9);    return ret;}double CST(int n, int m, double g){    init();    double U = m;    int S = 0, T = n + 1;    int i;    for (i = 1; i <= n; ++i) //<S,v,U>与<v,T,U+2g-dv>    {        add(S, i, U);        add(i, T, U + 2 * g - 1.0 * deg[i]);    }    for (i = 1; i <= m; ++i)//(u,v,1)    {        add(e[i].u, e[i].v, 1.0);        add(e[i].v, e[i].u, 1.0);    }    return dinic(S, T); //直接计算最小割值（最大流值）}int main(void){    int n, m, i;    while (~scanf("%d%d", &n, &m))    {        CLR(deg, 0);        for (i = 1; i <= m; ++i)        {            scanf("%d%d", &e[i].u, &e[i].v);            ++deg[e[i].u];            ++deg[e[i].v];        }        if (!m)        {            puts("1");            puts("1");        }        else        {            double L = 1.0 / n, R = m;            double ans = 1;            double dx = 1.0 / n / n;            while (fabs(R - L) >= dx)            {                double mid = (L + R) / 2.0;                double hg = (m * n * 1.0 - CST(n, m, mid)) / 2;                if (hg >= eps)                {                    L = mid;                    ans = mid;                }                else                    R = mid;            }            CST(n, m, ans);            vector<int>pos;            for (i = 1; i <= n; ++i)                if (~d[i])                    pos.push_back(i);            int sz = pos.size();            sort(pos.begin(), pos.end());            printf("%d\n", sz);            for (i = 0; i < sz; ++i)                printf("%d\n", pos[i]);        }    }    return 0;}