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;
const int MaxN = 100, MaxM = 200;
int n, m, all, Q;
int pre[MaxM + 5], last[MaxN + 5], other[MaxM + 5], col[MaxM + 5];
bool f[MaxN + 5][MaxN + 5][MaxM + 5], vis[MaxN + 5];
void Build(int x, int y, int c) {
  pre[++all] = last[x];
  last[x] = all;
  other[all] = y;
  col[all] = c;
}
void Dfs(int orig, int x, int c) {
  int ed = last[x], dr;
  while (ed != -1) {
    dr = other[ed];
    if (col[ed] == c && !vis[dr]) {
      vis[dr] = true;
      f[orig][dr][c] = true;
      Dfs(orig, dr, c);
    }
    ed = pre[ed];
  }
}
void Init() {
  all = -1;
  memset(last, -1, sizeof(last));
  for (int i = 1; i <= m; i++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    Build(u, v, c);
    Build(v, u, c);
  }
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; i++)
    for (int c = 1; c <= m; c++) {
      memset(vis, 0, sizeof(vis));
      vis[i] = true;
      Dfs(i, i, c);
    }
}
void Solve() {
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int u, v, cnt = 0;
    scanf("%d%d", &u, &v);
    for (int c = 1; c <= m; c++)
      if (f[u][v][c]) cnt++;
    printf("%d\n", cnt);
  }
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    Init();
    Solve();
  }
  return 0;
}
