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, i, now, pre, len, j, Q, x, y;
struct qq {
  int x, y, z;
  bool operator<(const qq &X) const { return z < X.z; }
} a[100100], b[100100];
int f[100100];
bool v[100100];
map<pair<int, int>, int> ma;
vector<int> G;
int fa(int o) {
  if (o == f[o]) return o;
  return f[o] = fa(f[o]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 0; i < m; ++i) {
    scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z);
  }
  sort(a, a + m);
  scanf("%d", &Q);
  for (i = 0; i < Q; ++i) {
    scanf("%d%d", &b[i].x, &b[i].y);
    if (b[i].x > b[i].y) swap(b[i].x, b[i].y);
  }
  now = 0;
  pre = 0;
  while (now < m) {
    while (a[pre].z == a[now].z) now++;
    G.clear();
    for (i = pre; i < now; ++i) {
      f[a[i].x] = a[i].x;
      f[a[i].y] = a[i].y;
      v[a[i].x] = v[a[i].y] = 1;
      G.push_back(a[i].x);
      G.push_back(a[i].y);
    }
    sort(G.begin(), G.end());
    len = unique(G.begin(), G.end()) - G.begin();
    for (i = pre; i < now; ++i) {
      f[fa(a[i].x)] = fa(a[i].y);
    }
    if (len * len <= Q) {
      for (i = 0; i < len; ++i) {
        for (j = i + 1; j < len; ++j) {
          if (fa(G[i]) == fa(G[j])) ma[make_pair(G[i], G[j])]++;
        }
      }
    } else {
      for (i = 0; i < Q; ++i) {
        if (v[b[i].x] && v[b[i].y]) {
          if (fa(b[i].x) == fa(b[i].y)) b[i].z++;
        }
      }
    }
    for (i = pre; i < now; ++i) v[a[i].x] = v[a[i].y] = 0;
    pre = now;
  }
  for (i = 0; i < Q; ++i) {
    if (ma.find(make_pair(b[i].x, b[i].y)) == ma.end()) {
      printf("%d\n", b[i].z);
    } else {
      printf("%d\n", ma[make_pair(b[i].x, b[i].y)] + b[i].z);
    }
  }
  return 0;
}
