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 = 100100;
pair<pair<int, int>, int> edges[MAXN];
unordered_set<int> col[MAXN];
unordered_map<int, int> par[MAXN];
unordered_map<long long, int> mp;
int find(int c, int x) {
  if (x == par[c][x]) return x;
  return par[c][x] = find(c, par[c][x]);
}
bool join(int c, int a, int b) {
  a = find(c, a);
  b = find(c, b);
  if (a == b) return false;
  par[c][a] = b;
  return true;
}
bool isconn(int c, int a, int b) { return find(c, a) == find(c, b); }
int main() {
  int N, M;
  scanf("%d %d", &N, &M);
  for (int i = 0; i < M; ++i) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    --a, --b, --c;
    edges[i] = make_pair(make_pair(a, b), c);
    col[a].insert(c);
    col[b].insert(c);
    par[c][a] = a;
    par[c][b] = b;
  }
  for (int i = 0; i < M; ++i) {
    join(edges[i].second, edges[i].first.first, edges[i].first.second);
  }
  int Q;
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    --a, --b;
    if (mp.count((long long)MAXN * a + b)) {
      printf("%d\n", mp[(long long)MAXN * a + b]);
      continue;
    }
    int cnt = 0;
    if (col[a].size() > col[b].size()) swap(a, b);
    for (int x : col[a]) {
      if (col[b].count(x)) {
        cnt += isconn(x, a, b);
      }
    }
    mp[(long long)MAXN * a + b] = cnt;
    mp[(long long)MAXN * b + a] = cnt;
    printf("%d\n", cnt);
  }
  return 0;
}
