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 read() {
  int a = 0, b = 1, c;
  while (((c = fgetc(stdin)) > '9' || c < '0') && c != '-')
    ;
  if (c == '-') b = -1, c = fgetc(stdin);
  while (c >= '0' && c <= '9') a = a * 10 + c - '0', c = fgetc(stdin);
  return a * b;
}
vector<pair<int, int> > e[100005];
vector<int> cc[100005], incc[100005];
int n, m, q, par[100005], ans[100005][450 + 5], ccrt[100005], bigind[100005];
inline int find(int x) { return x == par[x] ? x : par[x] = find(par[x]); }
inline void merge(int x, int y) { par[find(x)] = find(y); }
int main() {
  n = read();
  m = read();
  for (int i = 0; i < m; i++) {
    int u, v, c;
    u = read() - 1;
    v = read() - 1;
    c = read();
    e[c].push_back(make_pair(u, v));
  }
  int cnt_cc = 0;
  for (int i = 1; i <= m; i++) {
    vector<int> vec;
    for (int j = 0; j < ((int)(e[i]).size()); j++) {
      int u = e[i][j].first, v = e[i][j].second;
      par[u] = u;
      par[v] = v;
      ccrt[u] = ccrt[v] = 0;
      vec.push_back(u);
      vec.push_back(v);
    }
    sort(vec.begin(), vec.end());
    vec.erase(unique(vec.begin(), vec.end()), vec.end());
    for (int j = 0; j < ((int)(e[i]).size()); j++) {
      merge(e[i][j].first, e[i][j].second);
    }
    for (int j = 0; j < ((int)(vec).size()); j++) {
      int &cur = ccrt[find(vec[j])];
      if (!cur) cur = ++cnt_cc;
      cc[vec[j]].push_back(cur);
      incc[cur].push_back(vec[j]);
    }
  }
  int cnt_big = 0;
  for (int i = 0; i < n; i++) {
    if (((int)(cc[i]).size()) >= 450) {
      cnt_big++;
      bigind[i] = cnt_big;
      for (int j = 0; j < ((int)(cc[i]).size()); j++) {
        for (int k = 0; k < ((int)(incc[cc[i][j]]).size()); k++) {
          ans[incc[cc[i][j]][k]][cnt_big]++;
        }
      }
    }
  }
  q = read();
  for (int i = 0; i < q; i++) {
    int u, v;
    u = read() - 1;
    v = read() - 1;
    if (bigind[u] || bigind[v])
      printf("%d\n", bigind[u] ? ans[v][bigind[u]] : ans[u][bigind[v]]);
    else {
      int cnt = 0;
      for (int j = 0; j < ((int)(cc[u]).size()); j++) {
        auto ptr = lower_bound(cc[v].begin(), cc[v].end(), cc[u][j]);
        if (ptr != cc[v].end() && *ptr == cc[u][j]) cnt++;
      }
      printf("%d\n", cnt);
    }
  }
  return 0;
}
