liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many times, he is tired of such easy games and wants to play another game with Ikki.

liympanda has a magic circle and he puts it on a plane, there are n points on its boundary in circular border: 0, 1, 2, …, n − 1. Evil panda claims that he is connecting m pairs of points. To connect two points, liympanda either places the link entirely inside the circle or entirely outside the circle. Now liympanda tells Ikki no two links touch inside/outside the circle, except on the boundary. He wants Ikki to figure out whether this is possible…

Despaired at the minesweeping game just played, Ikki is totally at a loss, so he decides to write a program to help him.

The input contains exactly one test case.

In the test case there will be a line consisting of of two integers: n and m (n ≤ 1,000, m ≤ 500). The following m lines each contain two integers a_i and b_i, which denote the endpoints of the ith wire. Every point will have at most one link.

Output a line, either “panda is telling the truth...” or “the evil panda is lying again”.

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cmath>#include<string>#include<queue>#include<algorithm>#include<stack>#include<cstring>#include<vector>#include<list>#include<set>#include<map>using namespace std;#define ll long long#define pi (4*atan(1.0))#define eps 1e-14#define bug(x)  cout<<"bug"<<x<<endl;const int N=1e6+2,M=1e7+10,inf=1e9+10;const ll INF=1e18+10,mod=1e9+7;///   数组大小const int MAXNODE = 100005 * 2; //两倍struct TwoSat {    int n;    vector<int> g[MAXNODE];    int pre[MAXNODE], dfn[MAXNODE], dfs_clock, sccn, sccno[MAXNODE];    stack<int> S;    int mark[MAXNODE];    void init(int tot) {        n = tot * 2;        for (int i = 0; i < n; i+= 2) {            g[i].clear();            g[i^1].clear();        }    }    void add_Edge(int u, int v) {        g[u].push_back(v);        g[v].push_back(u);    }    void dfs_scc(int u) {        pre[u] = dfn[u] = ++dfs_clock;        S.push(u);        for (int i = 0; i < g[u].size(); i++) {            int v = g[u][i];            if (!pre[v]) {                dfs_scc(v);                dfn[u] = min(dfn[u], dfn[v]);            } else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);        }        if (pre[u] == dfn[u]) {            sccn++;            while (1) {                int x = S.top(); S.pop();                sccno[x] = sccn;                if (x == u) break;            }        }    }    void find_scc() {        dfs_clock = sccn = 0;        memset(sccno, 0, sizeof(sccno));        memset(pre, 0, sizeof(pre));        for (int i = 0; i < n; i++)            if (!pre[i]) dfs_scc(i);    }    bool solve() {        find_scc();        for (int i = 0; i < n; i += 2) {            if (sccno[i] == sccno[i + 1]) return false;            mark[i / 2] = (sccno[i] > sccno[i + 1]);        }        return true;    }} gao;int a[N],b[N];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        gao.init(n);        for(int i=1;i<=m;i++)        {            scanf("%d%d",&a[i],&b[i]);            if(a[i]>b[i])swap(a[i],b[i]);        }        for(int i=1;i<=m;i++)        {            for(int j=i+1;j<=m;j++)            {                if((a[i]<a[j]&&a[j]<b[i]&&b[j]>b[i])||(a[j]<a[i]&&a[i]<b[j]&&b[i]>b[j]))                {                    gao.add_Edge(2*i,2*j+1);                    gao.add_Edge(2*i+1,2*j);                }            }        }        if(gao.solve())printf("panda is telling the truth...\n");        else printf("the evil panda is lying again\n");    }    return 0;}