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.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.String.*;

public class Main {

    static int [][] p;
    
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
                                            //(new FileReader("input.in"));
        StringBuilder out = new StringBuilder();
        StringTokenizer tk;

        tk = new StringTokenizer(in.readLine());
        int n = parseInt(tk.nextToken()),m = parseInt(tk.nextToken());
        
        p = new int[m][n];
        for(int i=0; i<m; i++) 
            for(int j=0; j<n; j++) 
                p[i][j] = j;
        
        while(m-- > 0) {
            tk = new StringTokenizer(in.readLine());
            int a = parseInt(tk.nextToken())-1,b = parseInt(tk.nextToken())-1,c = parseInt(tk.nextToken())-1;
            a = find(a,c);
            b = find(b,c);
            
            if(a != b) {
                p[c][a] = b;
            }
        }
        
        int q = parseInt(in.readLine());
        
        while(q-- > 0) {
            tk = new StringTokenizer(in.readLine());
            int a = parseInt(tk.nextToken())-1,b = parseInt(tk.nextToken())-1;
            
            int ans = 0;
            for(int i=0; i<p.length; i++) {
                int x = find(a,i);
                int y = find(b,i);
                
                if(x==y) {
                    ans++;
                }
            }
            
            out.append(ans).append("\n");
        }
        
        System.out.print(out);
    }
    
    static int find(int u,int i) {
        if(p[i][u]==u)
            return u;
        return p[i][u] = find(p[i][u],i);
    }
}