John is the only priest in his town. September 1st is the John‘s busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples plan to get married on the blessed day. The i-th couple plan to hold their wedding from time S_i to time T_i. According to the traditions in the town, there must be a special ceremony on which the couple stand before the priest and accept blessings. The i-th couple need D_i minutes to finish this ceremony. Moreover, this ceremony must be either at the beginning or the ending of the wedding (i.e. it must be either from S_i to S_i + D_i, or from T_i - D_i to T_i). Could you tell John how to arrange his schedule so that he can present at every special ceremonies of the weddings.

Note that John can not be present at two weddings simultaneously.

The first line contains a integer N ( 1 ≤ N ≤ 1000). 
The next N lines contain the S_i, T_i and D_i. S_i and T_i are in the format of hh:mm.

The first line of output contains "YES" or "NO" indicating whether John can be present at every special ceremony. If it is "YES", output another N lines describing the staring time and finishing time of all the ceremonies.

#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 = 1010;const int MAXV = N * 2;const int MAXE = N * N * 4 * 2;struct edge{	int to, nxt;	edge() {}	edge(int _to, int _nxt): to(_to), nxt(_nxt) {}};edge E[MAXE];int head[MAXV], tot;int dfn[MAXV], low[MAXV], belong[MAXV], st[MAXV], top, ts, ins[MAXV], sc;int s[N], t[N], d[N];void init(){	CLR(head, -1);	tot = 0;	CLR(dfn, 0);	CLR(low, 0);	top = sc = 0;	CLR(ins, 0);	sc = 0;}inline void add(int s, int t){	E[tot] = edge(t, head[s]);	head[s] = tot++;}void Tarjan(int u){	low[u] = dfn[u] = ++ts;	st[top++] = u;	ins[u] = 1;	for (int i = head[u]; ~i; i = E[i].nxt)	{		int v = E[i].to;		if (!dfn[v])		{			Tarjan(v);			low[u] = min(low[u], low[v]);		}		else if (ins[v])			low[u] = min(low[u], dfn[v]);	}	if (low[u] == dfn[u])	{		++sc;		int v;		do		{			v = st[--top];			ins[v] = 0;			belong[v] = sc;		} while (u != v);	}}int main(void){	int n, i, j;	while (~scanf("%d", &n))	{		init();		for (i = 1; i <= n; ++i)		{			int h1, m1, h2, m2;			scanf(" %d:%d %d:%d %d", &h1, &m1, &h2, &m2, &d[i]);			s[i] = h1 * 60 + m1;			t[i] = h2 * 60 + m2;		}		for (i = 1; i <= n; ++i) //n*n*4*2		{			for (j = 1; j <= n; ++j)			{				if (i == j)					continue;				if (min(s[i] + d[i], s[j] + d[j]) > max(s[i], s[j])) //s-s				{					add(i, j + n);					add(j, i + n);				}				if (min(s[i] + d[i], t[j]) > max(s[i], t[j] - d[j])) //s-t				{					add(i, j);					add(j + n, i + n);				}				if (min(t[i], s[j] + d[j]) > max(t[i] - d[i], s[j])) //t-s				{					add(i + n, j + n);					add(j, i);				}				if (min(t[i], t[j]) > max(t[i] - d[i], t[j] - d[j])) //t-t				{					add(i + n, j);					add(j + n, i);				}			}		}		for (i = 1; i <= (n << 1); ++i)			if (!dfn[i])				Tarjan(i);		int flag = 1;		for (i = 1; i <= n; ++i)			if (belong[i] == belong[i + n])				flag = 0;		if (!flag)			puts("NO");		else		{			puts("YES");			for (i = 1; i <= n; ++i)			{				if (belong[i] < belong[i + n])					printf("%02d:%02d %02d:%02d\n", s[i] / 60, s[i] % 60, (s[i] + d[i]) / 60, (s[i] + d[i]) % 60);				else					printf("%02d:%02d %02d:%02d\n", (t[i] - d[i]) / 60, (t[i] - d[i]) % 60, (t[i]) / 60, (t[i]) % 60);			}		}	}	return 0;}