Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int d[110][110];
bool l[110][110][110];
bool vis[110];
vector<pair<int, int> > v[110];
void dfs(int st, int cur, int col) {
  if (!l[st][cur][col]) {
    d[st][cur]++;
    d[cur][st]++;
    l[st][cur][col] = true;
    l[cur][st][col] = true;
  } else
    return;
  for (int i = 0; i < v[cur].size(); i++)
    if (v[cur][i].second == col) dfs(st, v[cur][i].first, col);
}
void work(int x) {
  for (int i = 0; i < v[x].size(); i++) dfs(x, v[x][i].first, v[x][i].second);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0, x, y, z; i < m; i++) {
    scanf("%d %d %d", &x, &y, &z);
    v[x].push_back(make_pair(y, z));
    v[y].push_back(make_pair(x, z));
  }
  memset(vis, false, sizeof(vis));
  memset(l, false, sizeof(l));
  memset(d, 0, sizeof(d));
  scanf("%d", &q);
  for (int i = 0, u, v; i < q; i++) {
    scanf("%d %d", &u, &v);
    if (vis[u] || vis[v]) {
      printf("%d\n", d[u][v]);
    } else {
      vis[u] = true;
      work(u);
      printf("%d\n", d[u][v]);
    }
  }
  return 0;
}
