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, i, x, y, u, v, c, q, ans, j;
int ata[200005], size[200005], colsum[200005];
vector<int> col[200005];
pair<pair<int, int>, int> edges[200005];
unordered_map<int, int> dp[200005];
int atabul(int x) {
  if (ata[x] == -1) return x;
  return ata[x] = atabul(ata[x]);
}
int main() {
  scanf("%d %d", &n, &m);
  for (i = 1; i <= m; i++) {
    scanf("%d %d %d", &u, &v, &c);
    col[u].push_back(c);
    col[v].push_back(c);
    edges[i] = (make_pair(make_pair(u, v), c));
  }
  for (i = 1; i <= n; i++) {
    sort(col[i].begin(), col[i].end());
    col[i].resize(unique(col[i].begin(), col[i].end()) - col[i].begin());
    colsum[i + 1] = colsum[i] + col[i].size();
  }
  memset(ata, -1, sizeof ata);
  for (i = 0; i <= 200004; i++) size[i] = 1;
  for (i = 1; i <= m; i++) {
    u = edges[i].first.first;
    v = edges[i].first.second;
    c = edges[i].second;
    x = atabul(colsum[u] + lower_bound(col[u].begin(), col[u].end(), c) -
               col[u].begin());
    y = atabul(colsum[v] + lower_bound(col[v].begin(), col[v].end(), c) -
               col[v].begin());
    if (x == y) continue;
    if (size[x] < size[y]) swap(x, y);
    ata[y] = x;
    size[x] += size[y];
    size[y] = 0;
  }
  scanf("%d", &q);
  for (j = 1; j <= q; j++) {
    scanf("%d %d", &u, &v);
    ans = 0;
    if (col[u].size() > col[v].size()) swap(u, v);
    if (dp[u].find(v) != dp[u].end())
      ans = dp[u][v];
    else {
      for (i = 0; i < col[u].size(); i++) {
        c = col[u][i];
        if (binary_search(col[v].begin(), col[v].end(), c)) {
          x = colsum[v] + lower_bound(col[v].begin(), col[v].end(), c) -
              col[v].begin();
          if (atabul(x) == atabul(colsum[u] + i)) ans++;
        }
      }
      dp[u][v] = ans;
    }
    printf("%d\n", ans);
  }
}
