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>
const long long INF = 1e9;
const int P = 1e9 + 7;
const long long lINF = 1e18;
using namespace std;
int n, m, q;
unordered_map<int, int> prv[111111];
map<int, int> sz[111111];
map<pair<int, int>, int> answers;
vector<pair<int, int>> clrs[111111];
set<pair<int, int>> clrs2[111111];
int get(int v, int c) {
  if (!prv[v].count(c) || prv[v][c] == v) {
    return prv[v][c] = v;
  } else {
    return prv[v][c] = get(prv[v][c], c);
  }
}
void unite(int v1, int v2, int c) {
  int r1 = get(v1, c);
  int r2 = get(v2, c);
  if (!sz[v1][c]) {
    sz[v1][c] = 1;
  }
  if (!sz[v2][c]) {
    sz[v2][c] = 1;
  }
  if (sz[r1][c] > sz[r2][c]) {
    swap(r1, r2);
  }
  prv[r1][c] = r2;
  prv[r2][c] = r2;
  sz[r2][c] = sz[r1][c] + sz[r2][c];
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    unite(a, b, c);
  }
  for (int i = 1; i <= n; i++) {
    for (auto j : prv[i]) {
      if (j.second != 0) clrs[i].push_back({j.first, 0});
    }
  }
  for (int i = 1; i <= n; i++) {
    for (auto& j : clrs[i]) {
      clrs2[i].insert(make_pair(j.first, get(i, j.first)));
      j.second = get(i, j.first);
    }
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (answers.count({u, v})) {
      printf("%d\n", answers[{u, v}]);
      continue;
    }
    if (clrs[u].size() > clrs[v].size()) {
      swap(u, v);
    }
    int ans = 0;
    for (auto i : clrs[u]) {
      if (clrs2[v].count(i) && i.second != 0) {
        ans++;
      }
    }
    answers[{u, v}] = ans;
    answers[{v, u}] = ans;
    printf("%d\n", ans);
  }
  return 0;
}
