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;
int par[100005];
int r[100005];
void init() {
  for (int i = 0; i <= 100000; ++i) par[i] = i;
}
int getpar(int u) { return (par[u] == u ? u : (par[u] = getpar(par[u]))); }
void join(int u, int v) {
  int x = getpar(u);
  int y = getpar(v);
  if (x == y) return;
  if (r[x] != r[y]) {
    if (r[x] > r[y])
      par[y] = x;
    else
      par[x] = y;
  } else {
    r[x]++;
    par[y] = x;
  }
}
int N, M, Q;
int u, v, c;
map<int, int> color[100005];
map<pair<int, int>, int> ans;
int main() {
  scanf("%d%d", &N, &M);
  init();
  for (int i = 1; i <= M; ++i) {
    scanf("%d%d%d", &u, &v, &c);
    int a = 0, b = 0;
    if (color[u].count(c)) a = color[u][c];
    if (color[v].count(c)) b = color[v][c];
    color[u][c] = i;
    color[v][c] = i;
    if (a > 0) join(i, a);
    if (b > 0) join(i, b);
  }
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &u, &v);
    if (u > v) swap(u, v);
    if (ans.count(make_pair(u, v))) {
      printf("%d\n", ans[make_pair(u, v)]);
      continue;
    }
    int ret = 0;
    if (color[u].size() > color[v].size()) swap(u, v);
    for (map<int, int>::iterator uit = color[u].begin(); uit != color[u].end();
         ++uit) {
      map<int, int>::iterator vit = color[v].find(uit->first);
      if (vit == color[v].end()) continue;
      if (getpar(uit->second) == getpar(vit->second)) ++ret;
    }
    if (u > v) swap(u, v);
    ans[make_pair(u, v)] = ret;
    printf("%d\n", ret);
  }
  return 0;
}
