//// I'm lanjiangzhou
////C
//#include <stdio.h>
//#include <stdlib.h>
// I'm lanjiangzhou
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif//**************************CONSTANT***********************
#define INF 0x3f3f3f3f// aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//end
const int maxn =1010;
int head[maxn];
int vis[maxn];
int _count[maxn];//用来记录同一个点入队列的次数
int n,cnt;struct point{int v,next;//终点，指向下一条边int w;//权值
}edge[maxn*maxn];int dis[maxn];void CreateMap(int u,int v,int w){edge[cnt].v=v;edge[cnt].w=w;edge[cnt].next=head[u];head[u]=cnt++;
}int spfa(){memset(vis,0,sizeof(vis));memset(_count,0,sizeof(_count));for(int i=1;i<=n;i++){dis[i]=INF;}dis[1]=0;vis[1]=1;_count[1]++;queue<int> Q;Q.push(1);while(!Q.empty()){int tmp=Q.front();Q.pop();vis[tmp]=0;for(int i=head[tmp];i!=-1;i=edge[i].next){if(dis[tmp]+edge[i].w<dis[edge[i].v]){dis[edge[i].v]=dis[tmp]+edge[i].w;if(!vis[edge[i].v]){Q.push(edge[i].v);vis[edge[i].v]=1;if((++_count[edge[i].v])>n)return -1;//存在负权回路
                }}}}if(dis[n]==INF) return -2;return dis[n];//return 1;
}int main(){//int temp;//double l,u;int t;int x,y;int u,v,w;scanf("%d",&t);while(t--){memset(head,-1,sizeof(head));scanf("%d%d%d",&n,&x,&y);cnt=1;//int temp;for(int i=1;i<=x;i++){scanf("%d%d%d",&u,&v,&w);CreateMap(u,v,w);// for(int j=1;j<=m;j++){//scanf("%d",&temp);// CreateMap(i,j+n,-log(l*1.0/temp));// CreateMap(j+n,i,log(u*1.0/temp));// }
        }for(int i=1;i<=y;i++){scanf("%d%d%d",&u,&v,&w);CreateMap(v,u,-w);}printf("%d\n",spfa());//int flag=spfa();//if(flag) printf("%d\n",dis[n]);//else  if(dis[n]==INF) printf("-2\n");//else  if(!flag)   printf("-1\n");
    }return 0;
}

// I'm lanjiangzhou
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif//**************************CONSTANT***********************
#define INF 0x3f3f3f3f//aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//endconst int nmax = 1000+10;
const int emax = 23000+100;
int n;//一共有几个大营
int m;//已知从第i个大营到第j个大营至少有几个士兵，这些信息有m条
int c[nmax];//第i个大营最多有c[i]个士兵
int d[nmax];//d[0]=0;  d[1]=c[1];  d[2]=c[1]+c[2]....即d[i]为前i个大营容量总和
int ei;//边的序号
int dis[nmax];//从源点到个顶点最短路径长度（注意：源点为sn而不是s0）struct eg{int u,v,w;//边：起点，终点，权值
}edges[emax];void init(){ei=0;//除源点sn外，其他顶点的最短距离初始为inf// 则bellman_ford算法的第一重要循环n-1次for(int i=1;i<=n;i++) {dis[i]=INF;}dis[1]=0;dis[n]=0; //一下bellman_ford算法是以sn为源点的，所以dis[n]=0;
}bool bellman_ford(){//init();//bellman_ford()//算法的第一重循环要执行n-1次，n是网络中顶点的个数//本题中顶点个数n+1;for(int i=1;i<=n;i++){//   假设第k条边的起点是u，终点是v，以下循环考虑第k条边是否会使得源点v0到v的最短距离缩短，//  即判断dis[edges[k].u]+edges[k].w<dis[edges[k].v]是否成立for(int k=1;k<=ei;k++){int t=dis[edges[k].u]+edges[k].w;if(dis[edges[k].u]!=INF&&t<dis[edges[k].v]){dis[edges[k].v]=t;// printf("t=%d\n",t);
            }}}//以下是检查，若还有更新则说明存在无限循环的负值回路for(int k=1;k<=ei;k++){if(dis[edges[k].u]!=INF&&dis[edges[k].u]+edges[k].w<dis[edges[k].v]){return -1;}}if(dis[n]==INF) return -2;return dis[n];//return true;
}int main(){int t;scanf("%d",&t);while(t--){//init();int x,y;scanf("%d%d%d",&n,&x,&y);//init();int u,v,w;for(int i=1;i<=x;i++){scanf("%d%d%d",&u,&v,&w);edges[ei].u=u;edges[ei].v=v;edges[ei].w=w;ei++;}for(int j=1;j<=y;j++){scanf("%d%d%d",&u,&v,&w);edges[ei].u=v;edges[ei].v=u;edges[ei].w=-w;//构造边<v,u-1>  权值为-w;ei++;}init();printf("%d\n",bellman_ford());// int ans=bellman_ford();//if(ans) printf("-1\n");//else printf("%d\n",dis[n]);
    }return 0;
}