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


public class bfs_dfs {
	
	public static boolean[] vis;
	
	public static ArrayList<pair>[] v;
	
	@SuppressWarnings("unchecked")
	public static void main(String[] args)
	 {
		scanner in = new scanner();
		int n = in.nextInt();
		int m = in.nextInt();
		v = new ArrayList[n+1];
		for(int i = 0;i<=n;i++)
			v[i] = new ArrayList<>();
		for(int i = 0;i<m;i++)
		{
			int a = in.nextInt();
			int b = in.nextInt();
			int c = in.nextInt();
			v[a].add(new pair(b,c));
			v[b].add(new pair(a,c));
		}
		int q = in.nextInt();
		for(int i = 0;i<q;i++)
		{
			int str = in.nextInt();
			int dest = in.nextInt();
			int ans = 0;
			for(int j = 1;j<=m;j++)
			{
				 vis = new boolean[n+1];
				dfs(str,j);
				if(vis[dest]) ans++;
			}
			System.out.println(ans);
		}
	}
	public static void dfs(int str,int color)
	{
		vis[str] = true;
			for(pair i : v[str])
			{
				if( !vis[i.next] && i.color==color)
				{
					dfs(i.next,color);
				}
			}
	}
}
class pair
{
	int next;
	int color;
	public pair(int next,int color)
	{
		this.next = next;
		this.color = color;
	}
}
class scanner
{
	 BufferedReader br;
	 StringTokenizer st;
	
	public scanner()
	{
		br = new BufferedReader(new InputStreamReader(System.in));
		st = null;
	}
	public String next()
	{
		while(st==null || !st.hasMoreElements())
		{
			try
			{
				st = new StringTokenizer(br.readLine());
			}
			catch(IOException e){
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}
	public int nextInt()
	{
		return Integer.parseInt(next());
	}
	public long nextLong()
	{
    	return	Long.parseLong(next());
	}
}