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 MAX = 100005;
const int S = 300;
vector<int> adj[MAX], ver[MAX];
vector<pair<int, int> > edge[MAX];
unordered_map<int, int> mp[MAX];
int par[S + 50][MAX];
int find(int c, int v) {
  return (par[c][v] == -1 ? v : par[c][v] = find(c, par[c][v]));
}
bool merge(int c, int u, int v) {
  u = find(c, u);
  v = find(c, v);
  if (u == v) return false;
  par[c][u] = v;
  return true;
}
int pp[MAX], ch[MAX], sz;
void clear() {
  for (int i = 0; i < sz; i++) pp[ch[i]] = -1;
  sz = 0;
}
int find(int v) { return (pp[v] == -1 ? v : pp[v] = find(pp[v])); }
bool merge(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) return false;
  ch[sz++] = u;
  pp[u] = v;
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  memset(par, -1, sizeof(par));
  memset(pp, -1, sizeof(pp));
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    edge[w].push_back(make_pair(u, v));
    ver[w].push_back(u);
    ver[w].push_back(v);
  }
  for (int i = 0; i < MAX; i++) {
    sort(ver[i].begin(), ver[i].end());
    ver[i].resize(unique(ver[i].begin(), ver[i].end()) - ver[i].begin());
  }
  int t = 0;
  for (int i = 0; i < MAX; i++)
    if (edge[i].size() > S) {
      for (int j = 0; j < edge[i].size(); j++)
        merge(t, edge[i][j].first, edge[i][j].second);
      t++;
    } else {
      clear();
      for (int j = 0; j < edge[i].size(); j++)
        merge(edge[i][j].first, edge[i][j].second);
      for (int j = 0; j < ver[i].size(); j++)
        for (int k = j + 1; k < ver[i].size(); k++) {
          int u = ver[i][j];
          int v = ver[i][k];
          if (u > v) swap(u, v);
          if (find(u) == find(v)) mp[u][v]++;
        }
    }
  int q;
  cin >> q;
  while (q--) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    if (u > v) swap(u, v);
    int ans = mp[u][v];
    for (int i = 0; i < t; i++)
      if (find(i, u) == find(i, v)) ans++;
    cout << ans << "\n";
  }
  return 0;
}
