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, andW (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.

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
//#include <map>
using namespace std;

const int VM = 520;
const int EM = 5020;
const int INF = 0x3f3f3f3f;
//const int INF = 0x3f3f3f3f;

struct Edge{
    int u,v;
    int w;
}edge[EM<<1];

int N,M,W,cnt,dis[VM];

void addage(int cu,int cv,int cw){
    edge[cnt].u = cu;
    edge[cnt].v = cv;
    edge[cnt].w = cw;
    cnt++;
}

int Bellman(){
    int i,j,flag;
    for(i=0;i<N;i++){
        dis[i] = INF;
    }
    for(i=1;i<N;i++){
        flag = 1;
        for(j=0;j<cnt;j++){
            if(dis[edge[j].u] > dis[edge[j].v]+edge[j].w){
                dis[edge[j].u] = dis[edge[j].v]+edge[j].w;
                flag = 0;
            }
        }
        if(flag){
            break;
        }
    }
    for(i=0;i<cnt;i++){
        if(dis[edge[i].u] > dis[edge[i].v]+edge[i].w){
            return 1;
        }
    }

    return 0;
}

int main(){
    int t;
    int i;

    while(~scanf("%d",&t)){
        while(t--){
            scanf("%d%d%d",&N,&M,&W);
            cnt = 0;
            int u,v,w;
            for(i=0;i<M;i++){
                scanf("%d%d%d",&u,&v,&w);
                addage(u,v,w);
                addage(v,u,w);
            }
            for(i=0;i<W;i++){
                scanf("%d%d%d",&u,&v,&w);
                addage(u,v,-w);
            }
            if(Bellman()){
                printf("YES\n");
            }
            else{
                printf("NO\n");
            }
        }
    }

    return 0;
}