codeforces 505B——DFS——Mr. Kitayuta's Colorful Graph

Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the graph are numbered from 1 ton. Each edge, namely edge i, has a color c_i, connecting vertex a_i and b_i.

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

In the i-th query, he gives you two integers — u_i and v_i.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex u_i and vertex v_i 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 — a_i, b_i (1 ≤ a_i < b_i ≤ n) and c_i (1 ≤ c_i ≤ 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, (a_i, b_i, c_i) ≠ (a_j, b_j, c_j).

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 — u_i and v_i (1 ≤ u_i, v_i ≤ n). It is guaranteed that u_i ≠ v_i.

Output

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

Sample Input

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

Hint

Let's consider the first sample.

 The figure above shows the first sample.

• Vertex 1 and vertex 2 are connected by color 1 and 2.
• Vertex 3 and vertex 4 are connected by color 3.
• Vertex 1 and vertex 4 are not connected by any single color.

/*
题意：求从i到j有几条能通过的(颜色相同)的路径
多开一维保存颜色，暴力搜索
对节点i,j对不同颜色进行dfs
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
int map[110][110];
int vis[110];
int vis1[110][110];
int color[110][110][110];
vector<int> G[110];
int sum;

bool dfs(int u,int vv,int color1)
{
    if(!vis[u]){
        vis[u] = 1;
    if(u == vv) return true;
    for(int i = 0 ; i < G[u].size(); i++){
        int v = G[u][i];
        for(int j = 1 ; j <= vis1[u][v]; j++){
        if(color[u][v][j] == color1){
           if( dfs(v, vv, color1)) return true;
        }
        }
    }
    }
    return false ;
}

int main()
{
int n, m, k;
int a, b, c;
while(~scanf("%d %d", &n, &m)){
    memset(vis, 0, sizeof(vis1));
    for(int  i= 1; i <= m; i++){
        scanf("%d%d%d", &a, &b, &c);
        G[a].push_back(b);
        G[b].push_back(a);
        vis1[a][b]++;
        vis1[b][a]++;
        color[a][b][vis1[a][b]] = c;
        color[b][a][vis1[b][a]] = c;
    }
    scanf("%d", &k);
    int v1, v2;
    while(k--){
        sum = 0;
        scanf("%d%d", &v1, &v2);
        for(int i = 1; i <= m; i++){
            memset(vis, 0, sizeof(vis));
            if(dfs(v1, v2, i)){
                sum++;
            }
        }
        printf("%d\n", sum);
    }
}
  return 0;
}