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, q;
vector<pair<int, int> > edge[100001];
int par[100001];
vector<int> v[100001];
vector<int> chi[100001];
vector<pair<int, int> > res[100001];
map<pair<int, int>, int> mp;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int x, y, c;
    scanf("%d%d%d", &x, &y, &c);
    x--, y--, c--;
    edge[c].push_back(make_pair(x, y));
    v[c].push_back(x);
    v[c].push_back(y);
  }
  for (int i = 0; i < m; i++) {
    sort(v[i].begin(), v[i].end());
    vector<int> vt;
    for (int j = 0; j < v[i].size(); j++)
      if (!j || v[i][j] != v[i][j - 1]) vt.push_back(v[i][j]);
    for (int j = 0; j < vt.size(); j++) {
      chi[vt[j]].push_back(vt[j]);
      par[vt[j]] = vt[j];
    }
    for (int j = 0; j < edge[i].size(); j++) {
      int x = edge[i][j].first, y = edge[i][j].second;
      x = par[x], y = par[y];
      if (x == y) continue;
      if (chi[x].size() < chi[y].size()) swap(x, y);
      for (int k = 0; k < chi[y].size(); k++) {
        chi[x].push_back(chi[y][k]);
        par[chi[y][k]] = x;
      }
      chi[y].clear();
    }
    edge[i].clear();
    for (int j = 0; j < vt.size(); j++) {
      for (int k = 0; k < chi[vt[j]].size(); k++)
        res[chi[vt[j]][k]].push_back(make_pair(i, j));
      chi[vt[j]].clear();
    }
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    if (res[x].size() > res[y].size()) swap(x, y);
    if (mp.count(make_pair(x, y))) {
      printf("%d\n", mp[make_pair(x, y)]);
      continue;
    }
    int r = 0;
    for (int j = 0; j < res[x].size(); j++) {
      pair<int, int> p = res[x][j];
      if (*lower_bound(res[y].begin(), res[y].end(), p) == p) r++;
    }
    printf("%d\n", r);
    mp[make_pair(x, y)] = r;
  }
  return 0;
}
