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 + 10;
int n, m, q, color[N];
unordered_map<int, int> par[N], h[N], saved[N];
vector<pair<int, pair<int, int> > > edge;
void input() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v, c;
    scanf("%d %d %d", &u, &v, &c);
    u--, v--, c--;
    edge.push_back(make_pair(c, make_pair(u, v)));
  }
}
int findpar(int v, int c) {
  return par[v][c] = (par[v][c] == v ? v : findpar(par[v][c], c));
}
void merge(int u, int v, int c) {
  u = findpar(u, c);
  v = findpar(v, c);
  if (u == v) return;
  if (h[u][c] < h[v][c]) swap(v, u);
  par[v][c] = u;
  h[u][c] += h[v][c];
  return;
}
void dsu() {
  sort(edge.begin(), edge.end());
  for (int i = 0; i < m; i++) {
    int pt = i;
    while (edge[pt].first == edge[i].first && pt < m) {
      int u = edge[pt].second.first;
      int v = edge[pt].second.second;
      int c = edge[pt].first;
      if (par[u].count(c) == 0) color[u]++, h[u][c] = 1, par[u][c] = u;
      if (par[v].count(c) == 0) color[v]++, h[v][c] = 1, par[v][c] = v;
      merge(u, v, c);
      pt++;
    }
    i = pt - 1;
  }
}
int main() {
  input();
  dsu();
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--, y--;
    if (x > y) swap(x, y);
    if (saved[x].count(y) != 0)
      printf("%d \n", saved[x][y]);
    else {
      if (color[x] > color[y]) swap(x, y);
      int ans = 0;
      for (auto &it : par[x]) {
        int col = it.first;
        if (par[y].count(col) == 0) continue;
        int a = findpar(x, col);
        int b = findpar(y, col);
        if (a == b) ans++;
      }
      if (x > y) swap(x, y);
      saved[x][y] = ans;
      printf("%d \n", ans);
    }
  }
}
