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 maxn = 1e5 + 5;
const int LIMIT = 100;
vector<pair<int, int> > edges[maxn], query;
map<pair<int, int>, int> all;
set<int> colors;
int f[maxn];
int find(int a) { return f[a] == a ? a : f[a] = find(f[a]); }
vector<int> arr;
void solve(int c) {
  for (int i = 0; i < (int)edges[c].size(); i++) {
    int u = edges[c][i].first, v = edges[c][i].second;
    if (!f[u]) {
      f[u] = u;
      arr.push_back(u);
    }
    if (!f[v]) {
      f[v] = v;
      arr.push_back(v);
    }
    int u_ = find(u), v_ = find(v);
    if (u_ != v_) f[u_] = v_;
  }
  if ((int)arr.size() <= LIMIT) {
    sort(arr.begin(), arr.end());
    for (int i = 0; i < (int)arr.size(); i++)
      for (int j = i + 1; j < (int)arr.size(); j++) {
        int u = arr[i], v = arr[j];
        int u_ = find(u), v_ = find(v);
        if (u_ == v_ && all.find(make_pair(u, v)) != all.end())
          all[make_pair(u, v)]++;
      }
  } else {
    for (map<pair<int, int>, int>::iterator it = all.begin(); it != all.end();
         it++) {
      int u = (it->first).first, v = (it->first).second;
      int u_ = find(u), v_ = find(v);
      if (u_ == v_ && u_) (it->second)++;
    }
  }
  while (arr.size()) {
    f[arr.back()] = 0;
    arr.pop_back();
  }
  return;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    edges[c].push_back(make_pair(a, b));
    colors.insert(c);
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (u > v) swap(u, v);
    query.push_back(make_pair(u, v));
    if (u != v) all[make_pair(u, v)] = 0;
  }
  for (set<int>::iterator it = colors.begin(); it != colors.end(); it++)
    solve(*it);
  for (int i = 0; i < (int)query.size(); i++)
    printf("%d\n", query[i].first == query[i].second ? (int)colors.size()
                                                     : all[query[i]]);
  return 0;
}
