cf 505B

B. Mr. Kitayuta's Colorful Graph

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color ci, connecting vertex ai and bi.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers — ui and vi.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex uiand vertex vi directly or indirectly.

Input

The first line of the input contains space-separated two integers — n and m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers — ai, bi (1 ≤ ai < bi ≤ n) and ci (1 ≤ ci ≤ m). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if i ≠ j, (ai, bi, ci) ≠ (aj, bj, cj).

The next line contains a integer — q (1 ≤ q ≤ 100), denoting the number of the queries.

Then follows q lines, containing space-separated two integers — ui and vi (1 ≤ ui, vi ≤ n). It is guaranteed that ui ≠ vi.

Output

For each query, print the answer in a separate line.

Sample test(s)

input

4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4

output

2
1
0

input

5 7
1 5 1
2 5 1
3 5 1
4 5 1
1 2 2
2 3 2
3 4 2
5
1 5
5 1
2 5
1 5
1 4

output

1
1
1
1
2

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<stack>
using namespace std;
int n,m,dp[105][105][105],q;
int main()
{
      int a,b,c;
      memset(dp,0,sizeof(dp));
      scanf("%d%d",&n,&m);
      for(int i=0;i<m;i++)
      {
           scanf("%d%d%d",&a,&b,&c);
           dp[a][b][c]=1,dp[b][a][c]=1;
      }
      for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++)
                  for(int j=1;j<=n;j++)
                  {
                        for(int c=1;c<=m;c++)
                        {
                              if(dp[i][k][c]&&dp[k][j][c])
                                    dp[i][j][c]=1;
                        }
                  }
      scanf("%d",&q);
      while(q--)
      {
            int ans=0;
            scanf("%d%d",&a,&b);
            for(int i=1;i<=m;i++)
                  if(dp[a][b][i])
                        ans++;
            printf("%d\n",ans);
      }
      return 0;
}