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 size = 100010;
const int bound = 100;
int n, m, qq;
int res[size], arr[size], father[size];
vector<pair<int, int> > edge[size];
pair<int, int> q[size];
map<int, int> ans[size];
int find(int x) { return father[x] == -1 ? x : father[x] = find(father[x]); }
void merge(int x, int y) {
  int xx = find(x), yy = find(y);
  if (xx != yy) {
    father[xx] = yy;
  }
}
void solve() {
  for (int c = 1; c <= m; c++) {
    for (auto i : edge[c]) {
      merge(i.first, i.second);
    }
    if (edge[c].size() > bound) {
      for (int j = 0; j < qq; j++) {
        if (find(q[j].first) == find(q[j].second)) {
          ++res[j];
        }
      }
    } else {
      int num = 0;
      for (auto i : edge[c]) {
        arr[num++] = i.first;
        arr[num++] = i.second;
      }
      sort(arr, arr + num);
      num = unique(arr, arr + num) - arr;
      for (int u = 0; u < num; u++) {
        for (int v = u + 1; v < num; v++) {
          if (find(arr[u]) == find(arr[v])) {
            ++ans[arr[u]][arr[v]];
          }
        }
      }
    }
    for (auto i : edge[c]) {
      father[i.first] = father[i.second] = -1;
    }
  }
}
int main() {
  int a, b, c;
  memset(father, -1, sizeof(father));
  memset(res, 0, sizeof(res));
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    edge[c].push_back(make_pair(a, b));
  }
  cin >> qq;
  for (int i = 0; i < qq; i++) {
    cin >> q[i].first >> q[i].second;
  }
  solve();
  for (int i = 0; i < qq; i++) {
    cout << res[i] +
                max(ans[q[i].first][q[i].second], ans[q[i].second][q[i].first])
         << endl;
  }
  return 0;
}
