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 N = 100005, Q = 100005;
vector<pair<int, int> > cedge[N];
int u[Q], v[Q], ans[Q];
map<pair<int, int>, int> anss;
int stamp, vis[N], fa[N];
int find(int u) { return u == fa[u] ? u : fa[u] = find(fa[u]); }
void merge(int u, int v) { fa[u] = v; }
int n, m;
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);
    --a, --b, --c;
    cedge[c].push_back(make_pair(a, b));
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d%d", u + i, v + i);
    --u[i], --v[i];
    ans[i] = 0;
  }
  for (int i = 0; i < m; ++i) {
    if (cedge[i].size() == 0) {
      continue;
    }
    vector<int> vs;
    for (int j = 0; j < (int)cedge[i].size(); ++j) {
      vs.push_back(cedge[i][j].first);
      vs.push_back(cedge[i][j].second);
    }
    sort(vs.begin(), vs.end());
    vs.erase(unique(vs.begin(), vs.end()), vs.end());
    ++stamp;
    for (int j = 0; j < (int)vs.size(); ++j) {
      int u = vs[j];
      fa[u] = u;
      vis[u] = stamp;
    }
    for (int j = 0; j < (int)cedge[i].size(); ++j) {
      int u = cedge[i][j].first, v = cedge[i][j].second;
      if (find(u) != find(v)) {
        merge(find(u), find(v));
      }
    }
    if ((long long)vs.size() * vs.size() > n) {
      for (int i = 0; i < q; ++i) {
        if (vis[u[i]] == stamp && vis[v[i]] == stamp) {
          ans[i] += (find(u[i]) == find(v[i]));
        }
      }
    } else {
      for (int i = 0; i < (int)vs.size(); ++i) {
        for (int j = i + 1; j < (int)vs.size(); ++j) {
          if (find(vs[i]) == find(vs[j])) {
            ++anss[make_pair(vs[i], vs[j])];
            ++anss[make_pair(vs[j], vs[i])];
          }
        }
      }
    }
  }
  for (int i = 0; i < q; ++i) {
    printf("%d\n", ans[i] + anss[make_pair(u[i], v[i])]);
  }
  return 0;
}
