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 backT 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.

The second issue spfa -. - As for why to use this adjacency list is mainly to heavy side undirected path may be directed to cover the wormhole are not seeking the shortest path must be saved so that each edge

So adjacency table is the best choice because of the direct wave vector on just fine thing

#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
#define clr(a,b) memset(a,b,sizeof(a));
#define inf 0x3f3f3f3f
using namespace std;
int f,n,m,w,visit[505],index[505],dis[505];
struct node
{
	int to;
	int power;
};
vector <node> ma [505]; // save data to use the adjacency list ma [i] [j] .to i j is the starting point is the number of edges reaches the point is to use brute a direct two-dimensional array structure FIG stored gross thief 

int spfa()
{
	queue<int> q;
	 q.push (1); // start point is a single-source shortest 
	dis[1]=0;
	visit[1]=1;
	index[1]=1;
	int in=0;
	while(!q.empty())
	{
		in=q.front();
		q.pop();
		visit[in]=0;
		if(index[in]>n)
		{
			return 1;
		}
		for(int i=0;i<ma[in].size();i++)
		{
			int to=ma[in][i].to;
			if(dis[to]>dis[in]+ma[in][i].power)
			{
		        dis[to]=dis[in]+ma[in][i].power;
				if(!visit[to])
				{
					visit[to]=1;
					q.push(to);
					index[to]++;						
				}
			}
		}
	}
	return 0;
}

int main()
{
	ios::sync_with_stdio(false);
	cin>>f;
	while(f--)
	{
		cin>>n>>m>>w;
		for(int i=0;i<=504;i++)
		ma[i].clear();
		clr(visit,0)
		clr(index,0)
		clr(dis,inf)
		for(int i=1;i<=m;i++)
		{
			int s,e,p;
			node g;
			cin>>s>>e>>p;
			g.to=e;
			g.power=p;
			ma[s].push_back(g);
			g.to=s;
			ma[e].push_back(g);
			
		}
		for(int i=1;i<=w;i++)
		{
			int s,e,p;
			node g;
			cin>>s>>e>>p;
			g.to=e;
			g.power=-p;
			ma[s].push_back(g);
		}
		if(spfa()) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0; 
}