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 vis[101000], T, a[101000], b[101000], c[101000];
vector<int> v[101000], V[101000], col[101000];
set<pair<int, int> > S[101000];
set<pair<int, int> >::iterator it;
vector<pair<int, int> > Root[101000];
int root[101000];
map<pair<int, int>, int> mp;
int ans[101000];
int Fa(int x) {
  if (x == root[x]) return x;
  return root[x] = Fa(root[x]);
}
void doit(int id) {
  int x = a[id];
  int y = b[id];
  int co = c[id];
  if (vis[x] != T) {
    vis[x] = T;
    root[x] = x;
    V[co].push_back(x);
  }
  if (vis[y] != T) {
    vis[y] = T;
    root[y] = y;
    V[co].push_back(y);
  }
  int x1 = Fa(x);
  int y1 = Fa(y);
  if (x1 == y1) return;
  root[x1] = y1;
  return;
}
void Add(int co) {
  for (int i = 0; i < (int)V[co].size(); i++) {
    int x = V[co][i];
    int z = Fa(x);
    Root[x].push_back(pair<int, int>(co, z));
  }
}
int main() {
  int n, m, q, st, en;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) v[i].clear();
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
    v[a[i]].push_back(i);
    v[b[i]].push_back(i);
    col[c[i]].push_back(i);
  }
  T = 0;
  for (int i = 1; i <= m; i++)
    if (col[i].size()) {
      T++;
      for (int j = 0; j < (int)col[i].size(); j++) {
        int k = col[i][j];
        doit(k);
      }
      Add(i);
    }
  for (int i = 1; i <= n; i++) sort(Root[i].begin(), Root[i].end());
  scanf("%d", &q);
  mp.clear();
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &st, &en);
    if (st > en) swap(st, en);
    int xx = mp[pair<int, int>(st, en)];
    if (xx)
      printf("%d\n", ans[xx]);
    else {
      mp[pair<int, int>(st, en)] = i;
      if (Root[st].size() > Root[en].size()) swap(st, en);
      for (int j = 0; j < (int)Root[st].size(); j++) {
        pair<int, int> a = Root[st][j];
        xx =
            lower_bound(Root[en].begin(), Root[en].end(), a) - Root[en].begin();
        int zz = Root[en].size();
        if (xx != zz) {
          pair<int, int> b = Root[en][xx];
          if (a.first == b.first && a.second == b.second) ans[i]++;
        }
      }
      printf("%d\n", ans[i]);
    }
  }
  return 0;
}
