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 ui and 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.

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

1
 3
 5

#include <bits/stdc++.h>
using namespace std;
int n, m, q, a, b, c, s, t;
bool v[105];
vector<int> adj[105][105];
int dp[105][105][105];
int dfs(int par, int cur, int t, int cg) {
  v[cur] = true;
  if (cur == t) dp[cur][t][cg] = 1;
  if (dp[cur][t][cg] != 0) return dp[cur][t][cg];
  int ret = 0;
  for (int x : adj[cur][cg]) {
    if (x == par) continue;
    if (v[x]) continue;
    ret += dfs(cur, x, t, cg);
  }
  dp[cur][t][cg] = ret;
  return dp[cur][t][cg];
}
int main() {
  memset(dp, 0, sizeof dp);
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &a, &b, &c);
    adj[a][c].push_back(b);
    adj[b][c].push_back(a);
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &s, &t);
    int ans = 0;
    for (int j = 1; j <= m; j++) {
      memset(v, 0, sizeof v);
      ans += dfs(0, s, t, j);
    }
    printf("%d\n", ans);
  }
  return 0;
}
