TJU3214&HDU2290

/*
限制路径的最短路径问题，而且多次查询。顶点个数不多
使用floyd算法。限制路径，要注意排序。
*/

// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef long long ll;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)

const double Pi = acos(-1.0);
const double eps = 1e-6;
const int IntMAX = 0x7fffffff;
const double DoubleMAX = 1e307;

#define TMIN(x,y) ( x<y?x:y )
////////////////////////
//#define DEBUG

const ll Inf = 40000000;
int dp[210][210][210];
int mp[210][210];
int dx[210];
int N,M,Q;
struct node
{
	int C;
	int dx;
	friend bool operator<(node A,node B)
	{
		return A.C<B.C;
	}
};
node cops[210];
int T;

int main()
{
	read;
	write;
	int a,b,c;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d %d",&N,&M);
		for(int i=0;i<N;i++)
		{
			scanf("%d",&cops[i].C);
			cops[i].dx = i;
		}
		sort(cops,cops+N);
		for(int i=0;i<N;i++)
		{
			dx[cops[i].dx] = i+1;
		}

		for(int i=1;i<=N;i++)
		{
			for(int j=1;j<=N;j++)
			{
				mp[i][j] = (i==j?0:Inf);
			}
		}

		for(int i=0;i<M;i++)
		{
			scanf("%d %d %d",&a,&b,&c);
			
			mp[dx[a]][dx[b]] = mp[dx[b]][dx[a]] = min(mp[dx[a]][dx[b]],c);
		}
		

		for(int i=1;i<=N;i++)
		{
			for(int j=1;j<=N;j++)
			{
				dp[i][j][0] = mp[i][j];
			}
		}

		//
		for(int k=1;k<=N;k++)
		{
			for(int i=1;i<=N;i++)
			{
				for(int j=1;j<=N;j++)
				{
					dp[i][j][k] = dp[i][j][k-1];
				}
			}

			for(int i=1;i<=N;i++)
			{
				for(int j=1;j<=N;j++)
				{
					if( dp[i][j][k]>dp[i][k][k-1]+dp[k][j][k-1] )
					{
						 dp[i][j][k] = dp[i][k][k-1]+dp[k][j][k-1]; 
					}
				}
			}
		}

		//
		scanf("%d",&Q);
		while(Q--)
		{
			scanf("%d %d %d",&a,&b,&c);
			a = dx[a];
			b = dx[b];
			int ans = Inf;
			ans = min(ans,dp[a][b][0]);
			for(int i=N-1;i>=0;i--)
			{
				if( cops[i].C<=c )
				{
					ans = min(ans,dp[a][b][i+1]);
					break;
				}
			}
			
			if( ans==Inf )
				printf("-1\n");
			else
				printf("%d\n",ans);
		}
		printf("\n");
	}
	return 0;
}