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：

通过虫洞可以返回N年前，判断从某个点出发能不能回到从前，其实就是判断有没有负环。

题解：

Bellman-Ford算法，如果图中不存在负圈，那么最短路不会经过同一个顶点两次（最多通过V-1条边），把d[i]初始化为0，就可以判断所有的负环了。

#include
    
     #include
     
      #include
      
       using namespace std;const int maxn=505,INF=0x3f3f3f3f;int d[maxn];int n,m;int cnt;struct edge{    int u,v,cost;}es[maxn*maxn];bool find_negative_loop(){    memset(d,0,sizeof(d));    for(int i=0;i
       
        d[e.u]+e.cost)            {                    d[e.v]=d[e.u]+e.cost;                if(i==n-1)                    return true;            }        }    return false;}int main(){    int t;    cin>>t;    while(t--)    {        int q;        cin>>n>>m>>q;        cnt=0;        for(int i=0;i
        
         >a>>b>>c;//反向的边也要存储            es[cnt].u=es[cnt+1].v=a;            es[cnt].v=es[cnt+1].u=b;            es[cnt].cost=es[cnt+1].cost=c;            cnt+=2;        }        for(int i=0;i
         
          >es[cnt].u>>es[cnt].v>>es[cnt].cost;//添加负权边 es[cnt++].cost*=-1; } if(find_negative_loop()) cout<<"YES"<