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.awt.*;

public class Main implements Runnable {

	static BufferedReader br = new BufferedReader(new InputStreamReader(
			System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(
			System.out));
	static StringTokenizer st = new StringTokenizer("");

	public static String next() {
		try {
			while (!st.hasMoreTokens()) {
				String s = br.readLine();
				if (s == null)
					return null;
				st = new StringTokenizer(s);
			}
			return st.nextToken();
		} catch (Exception e) {
			return null;
		}
	}

	public static void main(String[] asda) throws Exception {
		new Thread(null, new Main(), "Main", 1 << 26).start();
	}

	public void run() {
		int N = Integer.parseInt(next());
		int M = Integer.parseInt(next());
		
		// g[x] contains y if x owe to y
		UnionFind [] g = new UnionFind[M];
		for (int i = 0; i < M; i++)
			g[i] = new UnionFind(N);
		
		while	(M-- > 0)	{
			int a = Integer.parseInt(next()) - 1;
			int b = Integer.parseInt(next()) - 1;
			int c = Integer.parseInt(next()) - 1;
			
			g[c].union(a, b);
		}

		int Q = Integer.parseInt(next());
		while	(Q-- > 0)	{
			int a = Integer.parseInt(next()) - 1;
			int b = Integer.parseInt(next()) - 1;
			
			int cnt = 0;
			for (int i = 0; i < g.length; i++)
				cnt += g[i].connected(a, b) ? 1 : 0;
			
			out.println(cnt);
		}
		//
		out.flush();
		System.exit(0);
	}
}


class UnionFind	{
	int [] dad;
	UnionFind(int N)	{
		dad = new int [N];
		for (int i = 0; i < N; i++)
			dad[i] = i;
	}
	int root(int x)	{
		if	( x == dad[x] )
			return	x;
		return	dad[x] = root( dad[x] );
	}
	
	void union(int a, int b)	{
		int x = root(a);
		int y = root(b);
		if	( x != y )
			dad[x] = y;
	}
	boolean connected(int a, int b)	{
		return	root(a) == root(b);
	}
}

