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;
unordered_map<int, int> parent[100005];
unordered_map<int, int> depth[100005];
unordered_map<int, int> memo[100005];
int Find(int node, int color) {
  if (parent[node][color] == node) return node;
  parent[node][color] = Find(parent[node][color], color);
  return parent[node][color];
}
void Union(int a, int b, int color) {
  if (!parent[a].count(color)) {
    parent[a][color] = a;
  }
  if (!parent[b].count(color)) {
    parent[b][color] = b;
  }
  int roota(Find(a, color));
  int rootb(Find(b, color));
  if (depth[roota][color] < depth[rootb][color]) swap(roota, rootb);
  if (depth[roota][color] == depth[rootb][color]) depth[roota][color]++;
  parent[rootb][color] = roota;
}
int solve(int a, int b) {
  if (memo[a].count(b)) return memo[a][b];
  int res(0);
  if (parent[a].size() > parent[b].size()) swap(a, b);
  unordered_map<int, int>::iterator it;
  for (it = parent[a].begin(); it != parent[a].end(); it++) {
    int color(it->first);
    if (!parent[b].count(color)) continue;
    if (Find(a, color) == Find(b, color)) res++;
  }
  memo[a][b] = memo[b][a] = res;
  return res;
}
int main() {
  int n, m, q;
  cin >> n >> m;
  for (int a, b, c, i = 0; i < m; i++) {
    scanf("%d%d%d", &a, &b, &c);
    a--;
    b--;
    Union(a, b, c);
  }
  cin >> q;
  for (int a, b, i = 0; i < q; i++) {
    scanf("%d%d", &a, &b);
    a--;
    b--;
    printf("%d\n", solve(a, b));
  }
  return 0;
}
