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 MAX_N = 100005;
unordered_map<int, int> parent[MAX_N], ans[MAX_N];
int rank_uf[MAX_N];
int n, m, query;
int findRoot(int x, int c) {
  return (parent[x][c] == x ? x : parent[x][c] = findRoot(parent[x][c], c));
}
void unionSet(int x, int y, int c) {
  if (parent[x].find(c) == parent[x].end()) parent[x][c] = x;
  if (parent[y].find(c) == parent[y].end()) parent[y][c] = y;
  int xx = findRoot(x, c);
  int yy = findRoot(y, c);
  if (xx == yy) return;
  if (rank_uf[xx] < rank_uf[yy])
    parent[xx][c] = yy;
  else {
    parent[yy][c] = xx;
    if (rank_uf[xx] == rank_uf[yy]) rank_uf[xx]++;
  }
}
int main() {
  scanf("%d %d", &n, &m);
  while (m--) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    if (a > b) swap(a, b);
    unionSet(a, b, c);
  }
  scanf("%d", &query);
  while (query--) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (parent[x].size() > parent[y].size()) swap(x, y);
    if (ans[x].find(y) == ans[x].end()) {
      int t = 0;
      for (unordered_map<int, int>::iterator it = parent[x].begin();
           it != parent[x].end(); it++) {
        int c = it->first;
        if (parent[y].find(c) == parent[y].end()) continue;
        int xx = findRoot(x, c);
        int yy = findRoot(y, c);
        if (xx == yy) t++;
      }
      ans[x][y] = t;
    }
    printf("%d\n", ans[x][y]);
  }
  return 0;
}
