Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Main{
	public static void main(String[] args)throws IOException{
		Escanner sc = new Escanner();
		n = sc.nextInt();
		int m = sc.nextInt();
		g = new ArrayList[n][n];
		for(int i=0; i<m; i++){
			int u = sc.nextInt()-1;
			int v = sc.nextInt()-1;
			int c = sc.nextInt();
			if(g[u][v]==null)
				g[u][v] = new ArrayList<Integer>();
			g[u][v].add(c);
			if(g[v][u]==null)
				g[v][u] = new ArrayList<Integer>();
			g[v][u].add(c);
		}
		int q = sc.nextInt();
		for(int i=0; i<q; i++){
			visited = new boolean[n];
			paths = 0;
			u = sc.nextInt()-1;
			v = sc.nextInt()-1;
			dfs(u, -1);
			System.out.println(paths);
		}
	}
	
	static class Escanner{
		BufferedReader in;
		StringTokenizer st;
		Escanner(){
			in = new BufferedReader(new InputStreamReader(System.in));
			st = new StringTokenizer("");
		}
		int nextInt()throws IOException{
			if(st.hasMoreTokens())
				return Integer.parseInt(st.nextToken());
			st = new StringTokenizer(in.readLine());
			return nextInt();
		}
	}

	static ArrayList<Integer> g[][];
	static int n;
	static int u;
	static int v;
	static int paths;
	static boolean[] visited;

	static void dfs(int i, int c){
		if(i==v){
			paths++;
			return;
		}
		visited[i] = true;
		for(int j=0; j<n; j++){
			if(g[i][j]!=null && !visited[j]){
				for(int k=0; k<g[i][j].size(); k++){
					if(c == -1)
						dfs(j, g[i][j].get(k));
					else if(g[i][j].get(k)==c)
						dfs(j, c);
					visited[j] = false;
				}
			}
		}
	}
}