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 = 100000 + 1;
vector<int> G[maxn];
map<int, int> ans[maxn];
map<int, int> father[maxn];
map<int, int> si[maxn];
int getfather(int a, int c) {
  auto &nf(father[a]);
  auto ft(nf.find(c));
  if (ft == nf.end()) return a;
  return ft->second == a ? a : ft->second = getfather(ft->second, c);
}
int n, m, q;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    int fa(getfather(a, c));
    int fb(getfather(b, c));
    int &ta(father[fa][c]);
    ta = fb;
    G[a].push_back(c);
    G[b].push_back(c);
  }
  for (int i = 1; i <= n; ++i) {
    sort(G[i].begin(), G[i].end());
    auto pt(unique(G[i].begin(), G[i].end()));
    G[i].resize(distance(G[i].begin(), pt));
  }
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (u > v) swap(u, v);
    if (G[u].size() > G[v].size()) swap(u, v);
    auto ft(ans[u].find(v));
    if (ft != ans[u].end())
      printf("%d\n", ft->second);
    else {
      int ts(0);
      for (int c : G[u])
        if (getfather(u, c) == getfather(v, c)) ts++;
      ans[u][v] = ts;
      printf("%d\n", ts);
    }
  }
  return 0;
}
