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 = 200010;
vector<int> bel[maxn];
vector<pair<int, int> > qry[maxn];
map<pair<int, int>, int> M;
map<pair<int, int>, vector<int> > edges;
int vis[maxn], ans[maxn];
void gao(int n) {
  int i, j;
  for (i = 0; i < bel[n].size(); i++) {
    j = bel[n][i];
    vis[j] = n;
  }
  for (i = 0; i < qry[n].size(); i++) {
    int v = qry[n][i].second;
    if (M.count(make_pair(n, v))) {
      ans[qry[n][i].first] = M[make_pair(n, v)];
      continue;
    }
    if (M.count(make_pair(v, n))) {
      ans[qry[n][i].first] = M[make_pair(v, n)];
      continue;
    }
    for (j = 0; j < bel[v].size(); j++)
      if (vis[bel[v][j]] == n) ans[qry[n][i].first]++;
    M[make_pair(n, v)] = ans[qry[n][i].first];
  }
}
int vv[maxn], col[maxn];
int cont[maxn], siz;
int in[maxn], block_cnt = 0;
void dfs(int n, int c) {
  if (in[n] == block_cnt) return;
  cont[++siz] = n;
  in[n] = block_cnt;
  for (int i = 0; i < edges[make_pair(n, c)].size(); i++) {
    int e = edges[make_pair(n, c)][i];
    if (!vis[e >> 1]) {
      vis[e >> 1] = 1;
      dfs(vv[e], c);
    }
  }
}
int ind[maxn];
int cmp(int u, int v) { return bel[u].size() > bel[v].size(); }
int main() {
  int n, m, i, j, k, q, u, v, e = 2;
  cin >> n >> m;
  while (m--) {
    cin >> u >> v >> j;
    if (!edges.count(make_pair(u, j))) {
      vector<int> tmp;
      edges[make_pair(u, j)] = tmp;
    }
    if (!edges.count(make_pair(v, j))) {
      vector<int> tmp;
      edges[make_pair(v, j)] = tmp;
    }
    edges[make_pair(u, j)].push_back(e), vv[e] = v, col[e] = j, e++;
    edges[make_pair(v, j)].push_back(e), vv[e] = u, col[e] = j, e++;
  }
  for (i = 1; i <= (e >> 1); i++) {
    u = vv[i * 2], v = vv[i * 2 + 1];
    if (!vis[i]) {
      block_cnt++;
      siz = 0;
      dfs(u, col[i * 2 + 1]);
      for (j = 1; j <= siz; j++) bel[cont[j]].push_back(block_cnt);
    }
  }
  for (i = 1; i <= n; i++) ind[i] = i;
  sort(ind + 1, ind + 1 + n, cmp);
  cin >> q;
  for (i = 1; i <= q; i++) {
    cin >> u >> v;
    if (bel[u].size() < bel[v].size()) swap(u, v);
    qry[u].push_back(make_pair(i, v));
  }
  memset(vis, 0, sizeof(vis));
  for (i = 1; i <= n; i++) gao(ind[i]);
  for (i = 1; i <= q; i++) cout << ans[i] << endl;
}
