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, x, y, c;
vector<int> vx[100005];
vector<int> v;
vector<pair<int, int> > e[100005];
int q;
int qx[100005], qy[100005];
int fa[100005];
int ans[100005];
int find(int x) {
  if (fa[x] != x) fa[x] = find(fa[x]);
  return fa[x];
}
map<pair<int, int>, int> mp;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int c;
    scanf("%d%d%d", &x, &y, &c);
    vx[c].push_back(x);
    vx[c].push_back(y);
    e[c].push_back(make_pair(x, y));
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &qx[i], &qy[i]);
  }
  for (int c = 1; c <= m; ++c) {
    v.clear();
    for (int i = 0; i < vx[c].size(); ++i) {
      v.push_back(vx[c][i]);
      fa[vx[c][i]] = vx[c][i];
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    for (int i = 0; i < e[c].size(); ++i) {
      if (find(e[c][i].first) != find(e[c][i].second))
        fa[find(e[c][i].first)] = find(e[c][i].second);
    }
    if (v.size() > 450) {
      for (int i = 1; i <= q; ++i) {
        if (!binary_search(v.begin(), v.end(), qx[i])) continue;
        if (!binary_search(v.begin(), v.end(), qy[i])) continue;
        if (find(qx[i]) == find(qy[i])) ++ans[i];
      }
    } else {
      for (int i = 0; i < v.size(); ++i)
        for (int j = 0; j < v.size(); ++j) {
          if (i == j) continue;
          if (find(v[i]) == find(v[j])) ++mp[make_pair(v[i], v[j])];
        }
    }
  }
  for (int i = 1; i <= q; ++i)
    printf("%d\n", ans[i] + mp[make_pair(qx[i], qy[i])]);
  return 0;
}
