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 = 100100;
int n, m;
int cnt[N];
int u[N], v[N], a[N], b[N], c[N], q;
int vis[N], A[N], len;
vector<int> V[N];
int fa[N], dep[N];
map<pair<int, int>, int> mp;
int ans[N];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int Union(int x, int y) {
  x = find(x), y = find(y);
  if (x == y) return 0;
  if (dep[x] < dep[y]) swap(x, y);
  if (dep[x] == dep[y]) dep[x]++;
  fa[y] = x;
  return 1;
}
int main() {
  ios ::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i] >> b[i] >> c[i];
    cnt[c[i]]++;
    V[c[i]].push_back(i);
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> u[i] >> v[i];
    if (u[i] > v[i]) swap(u[i], v[i]);
  }
  int blo = sqrt(m);
  int C = 0;
  for (int i = 1; i <= m; i++) {
    if (cnt[i] == 0) continue;
    C++;
    if (cnt[i] <= blo) {
      len = 0;
      for (auto e : V[i]) {
        fa[a[e]] = a[e];
        fa[b[e]] = b[e];
        dep[a[e]] = dep[b[e]] = 1;
        if (vis[a[e]] != i) {
          vis[a[e]] = i;
          A[++len] = a[e];
        }
        if (vis[b[e]] != i) {
          vis[b[e]] = i;
          A[++len] = b[e];
        }
      }
      for (auto e : V[i]) {
        Union(a[e], b[e]);
      }
      sort(A + 1, A + len + 1);
      for (int j = 1; j <= len; j++) {
        for (int k = j + 1; k <= len; k++) {
          if (find(A[j]) == find(A[k])) mp[{A[j], A[k]}]++;
        }
      }
    } else {
      for (int i = 1; i <= n; i++) fa[i] = i, dep[i] = 1;
      for (auto e : V[i]) {
        Union(a[e], b[e]);
      }
      for (int i = 1; i <= q; i++) {
        if (find(u[i]) == find(v[i])) ans[i]++;
      }
    }
  }
  for (int i = 1; i <= q; i++) {
    if (u[i] == v[i])
      cout << C << '\n';
    else
      cout << mp[{u[i], v[i]}] + ans[i] << '\n';
  }
  return 0;
}
