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 double pi = acos(-1);
int n, m, Q, tot, cnt;
int ans[110000], u[110000], v[110000], id[110000], num[110000], bo[110000],
    q[110000], f[110000];
vector<pair<int, int> > edge[110000];
vector<int> E[110000], D[110000];
void dfs(int x) {
  if (bo[x]) return;
  bo[x] = cnt;
  for (int i : E[x]) dfs(i);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (m); ++i) {
    int x, y, k;
    scanf("%d%d%d", &x, &y, &k);
    edge[k].push_back(make_pair(x, y));
    num[k]++;
  }
  scanf("%d", &Q);
  for (int i = (1); i <= (Q); ++i) {
    scanf("%d%d", &u[i], &v[i]);
    if (u[i] > v[i]) swap(u[i], v[i]);
  }
  int mm = sqrt(m);
  for (int i = (1); i <= (m); ++i)
    if (num[i] > mm) {
      for (int j = (0); j <= (num[i] - 1); ++j) {
        int x = edge[i][j].first, y = edge[i][j].second;
        E[x].push_back(y);
        E[y].push_back(x);
      }
      for (int j = (1); j <= (n); ++j)
        if (!bo[j]) {
          cnt++;
          dfs(j);
        }
      for (int i = (1); i <= (Q); ++i)
        if (bo[u[i]] == bo[v[i]]) ans[i]++;
      for (int j = (1); j <= (n); ++j) E[j].clear();
      memset(bo, 0, sizeof(bo));
    }
  for (int i = (1); i <= (m); ++i)
    if (num[i] <= mm) {
      tot = 0;
      for (int j = (0); j <= (num[i] - 1); ++j) {
        int x = edge[i][j].first, y = edge[i][j].second;
        q[++tot] = x;
        q[++tot] = y;
        E[x].push_back(y);
        E[y].push_back(x);
      }
      sort(q + 1, q + tot + 1);
      tot = unique(q + 1, q + tot + 1) - q - 1;
      for (int j = (1); j <= (tot); ++j)
        if (!bo[q[j]]) {
          cnt++;
          dfs(q[j]);
        }
      for (int j = (1); j <= (tot); ++j)
        for (int k = (j + 1); k <= (tot); ++k)
          if (bo[q[j]] == bo[q[k]]) {
            D[q[j]].push_back(q[k]);
          }
      for (int j = (1); j <= (tot); ++j) bo[q[j]] = 0, E[q[j]].clear();
    }
  for (int i = (1); i <= (Q); ++i) id[i] = i;
  sort(id + 1, id + Q + 1,
       [&](const int i, const int j) { return u[i] < u[j]; });
  int j = 1;
  for (int i = (1); i <= (n); ++i) {
    if (j <= Q && u[id[j]] != i) continue;
    for (int k : D[i]) f[k]++;
    while (j <= Q && u[id[j]] == i) {
      ans[id[j]] += f[v[id[j]]];
      j++;
    }
    for (int k : D[i]) f[k] = 0;
  }
  for (int i = (1); i <= (Q); ++i) printf("%d\n", ans[i]);
}
