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.*;
public class div286B
{
	BufferedReader in;
	PrintWriter ob;
	StringTokenizer st;
	int n,m;
	boolean color[][][],visited[];
	ArrayList<Integer> gr[];
	public static void main(String[] args) throws IOException {
		new div286B().run();
	}
	void run() throws IOException {
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solve();
		ob.flush();
	}
	@SuppressWarnings("unchecked")
	void solve() throws IOException {
		n=ni();
		m=ni();
		color=new boolean[n+1][n+1][m+1];
		gr=new ArrayList[n+1];
		for (int i=0; i<=n ; i++ ) {
			gr[i]=new ArrayList<Integer>();
		}
		for (int i=1; i<=m ; i++ ) {
			int u=ni();
			int v=ni();
			int w=ni();
			gr[u].add(v);
			gr[v].add(u);
			color[u][v][w]=true;
			color[v][u][w]=true;
		}
		int q=ni();
		while(q-->0) {
			int source=ni();
			int dest=ni();
			int ans=0;
			for(int k=1; k<=m ; k++)
			{
				visited=new boolean[n+1];
				ans+=dfs(source,dest , k);
			}
			ob.println(ans);
		}
	}
	int dfs(int u ,int d ,int c) {
		if(u==d) {
			return 1;
		}
		visited[u]=true;
		int ret=0;
		for (int v : gr[u] ) {
			if(color[u][v][c] && !visited[v])
				ret=Math.max(ret,dfs(v,d,c));
		}
		return ret;
	}
	int ni() throws IOException {
		return Integer.parseInt(nextToken());
	}
	String nextToken() throws IOException {
		if(st==null || !st.hasMoreTokens())
			st=new StringTokenizer(in.readLine());
		return st.nextToken();
	}
}