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 + 5;
const int SQN = 330;
int n, m;
int a[N];
int b[N];
int c[N];
int cnt[N];
int q;
int qu[N];
int qv[N];
int ans[N];
int parent[N];
int sz;
int buf[N];
vector<int> v[N];
map<int, int> mp[N];
int find(int node) {
  if (parent[node] == node) {
    return node;
  }
  return parent[node] = find(parent[node]);
}
void join(int a, int b) {
  int x = find(a);
  int y = find(b);
  if (x != y) {
    parent[y] = x;
    buf[++sz] = y;
  }
}
void solvebig(int color) {
  for (int i : v[color]) {
    join(a[i], b[i]);
  }
  for (int i = 1; i <= q; ++i) {
    if (find(qu[i]) == find(qv[i])) {
      ++ans[i];
    }
  }
}
int arr[SQN * 2];
int siz;
void solvesmall(int color) {
  for (int i : v[color]) {
    join(a[i], b[i]);
  }
  siz = 0;
  for (int i : v[color]) {
    arr[++siz] = a[i];
    arr[++siz] = b[i];
  }
  sort(arr + 1, arr + 1 + siz);
  siz = unique(arr + 1, arr + 1 + siz) - arr - 1;
  for (int i = 1; i < siz; ++i) {
    for (int j = i + 1; j <= siz; ++j) {
      if (find(arr[i]) == find(arr[j])) {
        ++mp[arr[i]][arr[j]];
      }
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d %d", a + i, b + i, c + i);
    ++cnt[c[i]];
    v[c[i]].emplace_back(i);
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d %d", qu + i, qv + i);
  }
  for (int i = 1; i <= n; ++i) {
    parent[i] = i;
  }
  sz = 0;
  for (int i = 1; i <= m; ++i) {
    if (cnt[i] >= SQN) {
      solvebig(i);
    } else {
      solvesmall(i);
    }
    for (int i = 1; i <= sz; ++i) {
      parent[buf[i]] = buf[i];
    }
    sz = 0;
  }
  for (int i = 1; i <= q; ++i) {
    printf("%d\n", ans[i] + mp[min(qu[i], qv[i])][max(qu[i], qv[i])]);
  }
}
