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 = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
int n, m, cnt, fa[N], a[N], b[N], c[N];
unordered_map<int, bool> mp[N];
map<pair<int, int>, int> ans;
vector<pair<int, int> > vec[N];
int getRoot(int x) { return fa[x] == x ? x : getRoot(fa[x]); }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
    vec[c[i]].push_back(make_pair(a[i], b[i]));
  }
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= m; i++) {
    map<int, int> ma;
    for (pair<int, int> t : vec[i]) {
      int x = getRoot(t.first), y = getRoot(t.second);
      if (x != y) fa[x] = y;
    }
    for (pair<int, int> t : vec[i]) {
      int x = getRoot(t.first), y = getRoot(t.second);
      if (ma.find(x) != ma.end()) {
        mp[t.first][ma[x]] = true;
      } else {
        cnt++;
        mp[t.first][cnt] = true;
        ma[x] = cnt;
      }
      if (ma.find(y) != ma.end()) {
        mp[t.second][ma[y]] = true;
      } else {
        cnt++;
        mp[t.second][cnt] = true;
        ma[y] = cnt;
      }
    }
    for (pair<int, int> t : vec[i]) {
      fa[t.first] = t.first;
      fa[t.second] = t.second;
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (mp[u].size() > mp[v].size()) swap(u, v);
    if (ans.find(make_pair(u, v)) != ans.end()) {
      printf("%d\n", ans[make_pair(u, v)]);
    } else {
      int num = 0;
      for (auto t : mp[u]) {
        if (mp[v].find(t.first) != mp[v].end()) num++;
      }
      printf("%d\n", num);
      ans[make_pair(u, v)] = num;
    }
  }
  return 0;
}
