Mr. Kitayuta has just bought an undirected graph with 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 ui and vertex vi directly or indirectly.

Input

The first line of the input contains space-separated two integers - n and m(2 ≤ n ≤ 105, 1 ≤ m ≤ 105), 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 ≤ 105), 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.

Examples

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

Note

Let's consider the first sample. 

<image> 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. 
#include <bits/stdc++.h>
using namespace std;
struct node {
  int p, cr;
};
vector<node> g[105];
int vis[105], crr[105];
int n, m;
int dp[105][105];
int cnt;
void dfs(int u, int cr, int ed) {
  if (u == ed) {
    if (!crr[cr]) {
      cnt++;
      crr[cr] = 1;
    }
    return;
  }
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].p;
    if (!vis[v] && (cr == 0 || cr == g[u][i].cr)) {
      vis[v] = 1;
      dfs(v, g[u][i].cr, ed);
      vis[v] = 0;
    }
  }
}
void init() {
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      cnt = 0;
      memset(vis, 0, sizeof(vis));
      memset(crr, 0, sizeof(crr));
      vis[i] = 1;
      dfs(i, 0, j);
      dp[i][j] = dp[j][i] = cnt;
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v, cr;
    scanf("%d %d %d", &u, &v, &cr);
    g[u].push_back((node){v, cr});
    g[v].push_back((node){u, cr});
  }
  init();
  int t;
  cin >> t;
  while (t--) {
    int u, v;
    scanf("%d %d", &u, &v);
    printf("%d\n", dp[u][v]);
  }
}
