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;
int n, m, q;
unordered_map<int, int> fa[100010], Size[100010], ans[100010];
int get(int u, int col) {
  return u == fa[u][col] ? u : fa[u][col] = get(fa[u][col], col);
}
void merge(int u, int v, int col) {
  if (!fa[u][col]) fa[u][col] = u, Size[u][col] = 1;
  if (!fa[v][col]) fa[v][col] = v, Size[v][col] = 1;
  int x = get(u, col), y = get(v, col);
  if (x == y) return;
  if (Size[x][col] > Size[y][col]) swap(x, y);
  fa[x][col] = y;
  Size[y][col] += Size[x][col];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    merge(u, v, w);
  }
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (fa[u].size() > fa[v].size()) swap(u, v);
    if (ans[u].count(v)) {
      printf("%d\n", ans[u][v]);
      continue;
    }
    int res = 0;
    for (pair<int, int> i : fa[u])
      res += (fa[v].count(i.first) && get(u, i.first) == get(v, i.first));
    printf("%d\n", ans[u][v] = res);
  }
  return 0;
}
