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;
unordered_map<int, int> h[100010];
map<pair<int, int>, int> f;
vector<int> g[100010], w[100010];
int a[100010], b[100010], q[100010], fa[100010], n, m, cas, t;
bool v[100010];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int getf(int x) {
  if (fa[x] != x) fa[x] = getf(fa[x]);
  return fa[x];
}
void ins(int x) {
  if (!v[x]) {
    v[x] = 1;
    q[++t] = x;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  int mx = 0;
  for (int i = 1; i <= m; i++) {
    int c;
    a[i] = read();
    b[i] = read();
    c = read();
    w[a[i]].push_back(c);
    w[b[i]].push_back(c);
    g[c].push_back(i);
    mx = max(mx, c);
  }
  for (int i = 1; i <= n; i++) {
    fa[i] = i;
    sort(w[i].begin(), w[i].end());
    w[i].erase(unique(w[i].begin(), w[i].end()), w[i].end());
  }
  for (int c = 1; c <= mx; c++) {
    t = 0;
    for (int j = 0; j < g[c].size(); j++) {
      int i = g[c][j];
      ins(a[i]);
      ins(b[i]);
      int x = getf(a[i]), y = getf(b[i]);
      if (x != y) fa[x] = y;
    }
    for (int i = 1; i <= t; i++) {
      int x = q[i];
      h[c][x] = getf(x);
      v[x] = 0;
    }
    for (int i = 1; i <= t; i++) fa[q[i]] = q[i];
  }
  scanf("%d", &cas);
  while (cas--) {
    int x, y, ans = 0;
    x = read();
    y = read();
    if (w[x].size() > w[y].size()) swap(x, y);
    pair<int, int> tt(x, y);
    if (f.find(tt) == f.end()) {
      for (int i = 0; i < w[x].size(); i++) {
        int c = w[x][i];
        if (h[c].find(y) != h[c].end() && h[c][x] == h[c][y]) ans++;
      }
      f[tt] = ans;
    } else
      ans = f[tt];
    printf("%d\n", ans);
  }
}
