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, f[100021], Q, ans[100021], block, vis[100021], sta[100021];
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
struct ques {
  int a, b;
} q[100021];
map<pair<int, int>, int> Map;
map<pair<int, int>, int> hs;
struct edge {
  int a, b, c;
  bool operator<(const edge& y) const { return c < y.c; }
} e[100021];
void unio(int a, int b) {
  a = find(a), b = find(b);
  if (a == b) return;
  f[a] = b;
}
int main() {
  scanf("%d%d", &n, &m);
  block = sqrt(n);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &e[i].a, &e[i].b, &e[i].c);
  sort(e + 1, e + 1 + m);
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    scanf("%d%d", &q[i].a, &q[i].b);
    hs[make_pair(q[i].a, q[i].b)]++;
    hs[make_pair(q[i].b, q[i].a)]++;
  }
  for (int i = 1; i <= n; i++) f[i] = i;
  for (int x, j, i = 1; i <= m; i = j) {
    for (j = i + 1; j <= m; j++)
      if (e[j].c != e[j - 1].c) break;
    int cnt = 0;
    for (int k = i; k < j; k++) {
      unio(e[k].a, e[k].b);
      if (!vis[e[k].a]) sta[++cnt] = e[k].a;
      if (!vis[e[k].b]) sta[++cnt] = e[k].b;
      vis[e[k].a] = vis[e[k].b] = 1;
    }
    if (false) {
      for (int t = 1; t <= cnt; t++) {
        for (int k = t + 1; k <= cnt; k++) {
          if (hs[make_pair(sta[t], sta[k])])
            Map[make_pair(sta[t], sta[k])]++, Map[make_pair(sta[k], sta[t])]++;
        }
      }
    } else {
      for (int k = 1; k <= Q; k++) {
        if (find(q[k].a) == find(q[k].b)) {
          ans[k]++;
        }
      }
    }
    for (int k = 1; k <= n; k++) f[k] = k, vis[k] = 0;
  }
  for (int i = 1; i <= Q; i++)
    printf("%d\n", ans[i] + Map[make_pair(q[i].a, q[i].b)]);
  return 0;
}
