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. 
import java.io.*;
import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner s = new Scanner(System.in);
    int n = s.nextInt(), m = s.nextInt();
    Edge[] e = new Edge[m];
    for (int i = 0; i < m; i++) {
      e[i] = new Edge(s.nextInt() - 1, s.nextInt() - 1, s.nextInt());
    }
    int q = s.nextInt();
    Edge[] ans = new Edge[q];
    for (int i = 0; i < q; i++) {
      ans[i] = new Edge(s.nextInt() - 1, s.nextInt() - 1, 0);
    }
    int[] f = new int[n];
    int[] rank = new int[n];
    for (int c = 1; c <= m; c++) {
      for (int i = 0; i < n; i++) {
        f[i] = i;
        rank[i] = 0;
      }
      for (int i = 0; i < m; i++) {
        if (e[i].c == c) {
          union(f, rank, e[i].a, e[i].b);
        }
      }
      for (int i = 0; i < q; i++) {
        if (getFather(f, ans[i].a) == getFather(f, ans[i].b)) {
          ans[i].c++;
        }
      }
    }
    for (int i = 0; i < q; i++) {
      System.out.println(ans[i].c);
    }
  }

  public static int getFather(int[] f, int u) {
    if (f[u] == u) {
      return u;
    } else {
      int v = getFather(f, f[u]);
      f[u] = v;
      return v;
    }
  }

  public static void union(int[] f, int[] rank, int a, int b) {
    int fa = getFather(f, a), fb = getFather(f, b);
    if (fa != fb) {
      if (rank[fa] < rank[fb]) {
        f[fa] = fb; 
      } else if (rank[fa] > rank[fb]) {
        f[fb] = fa;
      } else {
        rank[fa]++;
        f[fb] = fa;
      }
    }
  }

  public static class Edge {
    int a, b, c;
    public Edge(int a, int b, int c) {
      this.a = a;
      this.b = b;
      this.c = c;
    }
  }
}