While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ’s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1.. F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).

Sample Input

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

题意：

一个人能否从一个起点出发， 走回来时间会不会倒流。

因为此题存在负权边的问题， 所以迪杰斯特拉算法行不通， 所以需要用Bellman-Ford来求， 但这个题目的意思并不是让你求最短路径， 而是转换成了是否会有负权环的问题。

所以在用完Bellman-Ford后， 利用负权边的判断来给出题目答案。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;
const int maxn=5505;
const int INF=0x3f3f3f3f;
int f;
int n,m,w;
struct edge
{
    int s;
    int len;
    int e;
};
edge e[maxn];
int d[maxn];
int num;
//Bellman-Ford算法
bool Bellman (int s)
{
    for (int i=1;i<=n;i++)
    {
        if(i==s)
            d[i]=0;
        else
            d[i]=INF;
    }
    for (int i=1;i<n;i++)
    {
        for (int j=0;j<num;j++)
        {
            if(d[e[j].e]>d[e[j].s]+e[j].len)
                d[e[j].e]=d[e[j].s]+e[j].len;
        }
    }
    //题目的关键所在
    for (int i=0;i<num;i++)
        if(d[e[i].e]>d[e[i].s]+e[i].len)
              return true;
    return false;
}
int main()
{
    scanf("%d",&f);
    while (f--)
    {
        num=0;
        scanf("%d%d%d",&n,&m,&w);
        for (int i=0;i<m;i++)
        {
           int x,y,len;
           scanf("%d%d%d",&x,&y,&len);
           e[num].s=x;
           e[num].len=len;
           e[num++].e=y;
           e[num].s=y;
           e[num].len=len;
           e[num++].e=x;

        }
        for (int i=0;i<w;i++)
        {
            int x,y,len;
            scanf("%d%d%d",&x,&y,&len);
            e[num].s=x;
            e[num].len=-len;
            e[num++].e=y;
        }
       if(Bellman(1))
          printf("YES\n");
       else
          printf("NO\n");
    }
    return 0;
}