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[200005];
map<pair<int, int>, int> cnt;
int n, m, q, qx[200005], qy[200005], par[200005], ans[200005];
inline int find(int x) { return x == par[x] ? x : par[x] = find(par[x]); }
inline void merge(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  par[y] = x;
}
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));
  }
  q = read();
  for (int i = 0; i < q; i++) {
    qx[i] = read() - 1;
    qy[i] = read() - 1;
    if (qx[i] > qy[i]) swap(qx[i], qy[i]);
  }
  for (int i = 1; i <= m; i++) {
    vector<int> vec;
    for (int j = 0; j < ((int)(e[i]).size()); j++) {
      vec.push_back(e[i][j].first);
      vec.push_back(e[i][j].second);
      par[e[i][j].first] = e[i][j].first;
      par[e[i][j].second] = e[i][j].second;
    }
    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);
    }
    if (((int)(vec).size()) <= 700) {
      for (int j = 0; j < ((int)(vec).size()); j++) {
        for (int k = j + 1; k < ((int)(vec).size()); k++) {
          if (find(vec[j]) == find(vec[k])) {
            cnt[make_pair(vec[j], vec[k])]++;
          }
        }
      }
    } else {
      for (int j = 0; j < q; j++) {
        if (*lower_bound(vec.begin(), vec.end(), qx[j]) != qx[j]) continue;
        if (*lower_bound(vec.begin(), vec.end(), qy[j]) != qy[j]) continue;
        if (find(qx[j]) == find(qy[j])) {
          ans[j]++;
        }
      }
    }
  }
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i] + cnt[make_pair(qx[i], qy[i])]);
  }
  return 0;
}
