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 = 100005;
unordered_map<int, int> pr[N], cnt[N];
unordered_map<int, unordered_map<int, int>> mp;
set<int> col[N];
vector<int> G[2 * N];
int n, m, q, vis[N], id, fid = 1;
int find(int c, int r) {
  int x = pr[c][r];
  if (x == 0) return pr[c][r] = r;
  if (x == r)
    return r;
  else
    return pr[c][r] = find(c, x);
}
void dfs(int root, int node, int c) {
  vis[node] = id;
  cnt[root][node]++;
  int h = mp[node][c];
  auto it = G[h].begin();
  auto pt = G[h].end();
  while (it != pt) {
    int go = *it;
    it++;
    if (vis[go] == id) continue;
    dfs(root, go, c);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    col[a].insert(c);
    col[b].insert(c);
    if (mp[a][c] == 0) {
      mp[a][c] = fid++;
    }
    if (mp[b][c] == 0) {
      mp[b][c] = fid++;
    }
    G[mp[a][c]].push_back(b);
    G[mp[b][c]].push_back(a);
    int u = find(c, a);
    int v = find(c, b);
    if (u != v) {
      pr[c][u] = pr[c][v];
    }
  }
  int sq = 100;
  for (int i = 1; i <= n; i++) {
    if (col[i].size() < sq) continue;
    for (int j : col[i]) {
      ++id;
      dfs(i, i, j);
    }
  }
  scanf("%d", &q);
  while (q--) {
    int a, b, ans = 0;
    scanf("%d %d", &a, &b);
    if (col[a].size() > col[b].size()) swap(a, b);
    if (col[a].size() < sq) {
      for (int i : col[a])
        if (find(i, a) == find(i, b)) ans++;
    } else {
      ans = cnt[a][b];
    }
    printf("%d\n", ans);
  }
}
