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 = 1e5 + 10;
tuple<int, int, int> E[MAX_N];
map<pair<int, int>, int> memo;
unordered_map<int, int> f[MAX_N], o[MAX_N];
int N, M, Q;
int getf(int i, int c) { return f[i][c] == i ? i : f[i][c] = getf(f[i][c], c); }
void mergef(int c, int i, int j) {
  i = getf(i, c), j = getf(j, c);
  if (i == j) return;
  if (o[i][c] == o[j][c]) {
    f[j][c] = i;
    o[i][c]++;
  } else {
    if (o[i][c] < o[j][c]) swap(i, j);
    f[j][c] = i;
  }
}
int main() {
  int u, v, c;
  scanf("%d%d", &N, &M);
  for (int i = 0; i < M; i++) {
    scanf("%d%d%d", &u, &v, &c);
    f[u][c] = u, o[u][c] = 0;
    f[v][c] = v, o[v][c] = 0;
    E[i] = make_tuple(u, v, c);
  }
  for (int i = 0; i < M; i++) {
    tie(u, v, c) = E[i];
    mergef(c, u, v);
  }
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &u, &v);
    if (f[u].size() > f[v].size()) swap(u, v);
    if (!memo.count({u, v}))
      for (auto it : f[u])
        if (f[v].find(it.first) != f[v].end() &&
            getf(v, it.first) == getf(u, it.first))
          memo[{u, v}] += 1;
    printf("%d\n", memo[{u, v}]);
  }
  return 0;
}
