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 maxn = 100007;
int n, m, q;
unordered_map<int, int> fa[maxn], ans[maxn];
int find_set(int v, int c) {
  if (fa[v][c] == v) return v;
  return fa[v][c] = find_set(fa[v][c], c);
}
void union_set(int x, int y, int c) {
  if (fa[x].find(c) == fa[x].end()) fa[x][c] = x;
  if (fa[y].find(c) == fa[y].end()) fa[y][c] = y;
  int fx = find_set(x, c), fy = find_set(y, c);
  if (fx != fy) fa[fx][c] = fy;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y, c;
    scanf("%d%d%d", &x, &y, &c);
    union_set(x, y, c);
  }
  scanf("%d", &q);
  while (q--) {
    int x, y, Ans = 0;
    scanf("%d%d", &x, &y);
    if (ans[x].find(y) != ans[x].end()) {
      printf("%d\n", ans[x][y]);
      continue;
    }
    int s1 = fa[x].size(), s2 = fa[y].size();
    if (s1 > s2) swap(x, y);
    for (auto u : fa[x]) {
      if (fa[y].find(u.first) != fa[y].end() &&
          find_set(x, u.first) == find_set(y, u.first))
        Ans++;
    }
    ans[x][y] = ans[y][x] = Ans;
    printf("%d\n", Ans);
  }
  return 0;
}
