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;
inline void read(int &x) {
  short negative = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') negative = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  x *= negative;
}
inline void write(long long x) {
  static long long sta[35];
  long long top = 0;
  do {
    sta[top++] = x % 10, x /= 10;
  } while (x);
  while (top) putchar(sta[--top] + 48);
  puts("");
}
bool prime(int n) {
  if (n == 0 || n == 1) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
vector<int> v[111][111];
bool use[111];
void dfs(int n, int c) {
  use[n] = true;
  for (int i = 0; i < (int)(v[n][c].size()); i++)
    if (!use[v[n][c][i]]) dfs(v[n][c][i], c);
}
int main() {
  int n, m;
  read(n);
  read(m);
  for (int i = 0; i < (int)(m); i++) {
    int a, b, c;
    read(a);
    read(b);
    read(c);
    v[a][c].push_back(b);
    v[b][c].push_back(a);
  }
  int q;
  read(q);
  for (int i = 0; i < (int)(q); i++) {
    int a, b, ans = 0;
    read(a);
    read(b);
    for (int j = 1; j <= 100; j++) {
      for (int k = 1; k <= 100; k++) use[k] = 0;
      dfs(a, j);
      if (use[b]) ans++;
    }
    write(ans);
  }
  return 0;
}
