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.util.*;
import java.io.*;
import java.math.*;
public class Euler {
	
	static class pair {
	  
	  int next;
	  int col;
	  public pair(int next, int col) {
		this.next = next;
		this.col = col;  
	  }	
		
	}
	static ArrayList<pair>[] arrayList;
	static boolean[] visited;
	public static void main(String[] args){
		
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		
		arrayList = new ArrayList[n + 1];
		for (int i = 1; i <= n; i++) arrayList[i] = new ArrayList<>();
		
		for (int i = 0; i < m; i++) {
			
			int a = in.nextInt();
			int b = in.nextInt();
			int c = in.nextInt();
			arrayList[a].add(new pair(b, c));
			arrayList[b].add(new pair(a, c));
			
		}
		
		int q = in.nextInt();
		while (q-- > 0) {
			int a = in.nextInt();
			int b = in.nextInt();
			int count = 0;
			for (int i = 1; i <= m; i++) {
				visited = new boolean[n + 1];
				dfs(a, i);
				if (visited[b]) count++;
			}
			System.out.println(count);
		}
		
	}
	
	public static void dfs(int node, int color) {
		visited[node] = true;
		for (pair pp : arrayList[node]) {
			if (!visited[pp.next]) {
				if (pp.col == color) dfs(pp.next, color);
			}
		}
	}
	
	
	
	
	
}


