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.ArrayList;
import java.util.Scanner;
import java.util.Vector;

public class Algorithm {

	public static void main(String[] args) {

		Algorithm a = new Algorithm();
	}

	Scanner scan;

	public Algorithm() {
		scan = new Scanner(System.in);
		readData();
	}
	int[][][] g;
	int n, m;
	int count = 0;
	boolean[][] vis;
	int dest;
	public void readData() {
		
		n = scan.nextInt();
		m = scan.nextInt();
		g = new int[n + 1][n + 1][m + 1];
		
		for (int k = 0; k < m; k++){
			int x = scan.nextInt(), y = scan.nextInt(), c = scan.nextInt();
			g[x][y][c] = 1;
			g[y][x][c] = 1;
		}
		
		int q = scan.nextInt();
		for (int k = 0; k < q; k++){
			int x = scan.nextInt(), y = scan.nextInt();
			vis = new boolean[n + 1][m + 1];
			dest = y;
			count = 0;
			for (int i = 1; i <= m; i++){
				if (dfs(x, i))
					count++;
			}
			System.out.println(count);
		}
	}	
	
	public boolean dfs(int a, int c){
//		System.out.println(a + " " + c);
		if (dest == a)
			return true;
		
		vis[a][c] = true;
		
		for (int k = 1; k <= n; k++){
			if (g[a][k][c] == 1 && !vis[k][c] && dfs(k, c))
				return true;
		}
		
		return false;
	}
	
}
