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 NICO = 100000 + 10;
int n, m, q;
int qu[NICO], qv[NICO];
int a[NICO], b[NICO], c[NICO];
vector<pair<int, int> > edge[NICO];
vector<int> room;
map<pair<int, int>, int> ans;
int res[NICO];
int par[NICO], id[NICO];
int find(int x) { return par[x] == -1 ? x : par[x] = find(par[x]); }
void unite(int x, int y) {
  int p1 = find(x);
  int p2 = find(y);
  if (p1 != p2) {
    par[p1] = p2;
  }
}
int main() {
  for (int i = 1; i <= n; i++) {
    par[i] = -1;
  }
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    edge[c].push_back(make_pair(a, b));
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%d %d", &qu[i], &qv[i]);
  }
  for (int i = 1; i <= m; i++) {
    room.clear();
    for (int j = 0; j < edge[i].size(); j++) {
      room.push_back(edge[i][j].first);
      room.push_back(edge[i][j].second);
    }
    sort(room.begin(), room.end());
    room.erase(unique(room.begin(), room.end()), room.end());
    for (int j = 0; j < room.size(); j++) {
      par[room[j]] = -1;
      id[room[j]] = i;
    }
    for (int j = 0; j < edge[i].size(); j++) {
      unite(edge[i][j].first, edge[i][j].second);
    }
    if ((long long)room.size() * (long long)room.size() > n) {
      for (int j = 1; j <= q; j++) {
        int u = qu[j], v = qv[j];
        if (id[u] == i && id[v] == i && find(u) == find(v)) {
          res[j]++;
        }
      }
    } else {
      for (int j = 0; j < room.size(); j++) {
        for (int k = j + 1; k < room.size(); k++) {
          if (find(room[j]) == find(room[k])) {
            ans[make_pair(room[j], room[k])]++;
            ans[make_pair(room[k], room[j])]++;
          }
        }
      }
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[make_pair(qu[i], qv[i])] + res[i]);
  }
}
