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.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;

public class ColorfulGraph {

	static boolean[] visited;
	static ArrayList<Edge>[] graph;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n, m;
		n = in.nextInt();
		m = in.nextInt();
		graph = new ArrayList[n];
		
		for(int i = 0; i < m; ++i) {
			int a = in.nextInt()-1;
			int b = in.nextInt()-1;
			int c = in.nextInt();
			if(graph[a] == null) graph[a] = new ArrayList<Edge>();
			if(graph[b] == null) graph[b] = new ArrayList<Edge>();
			graph[a].add(new Edge(b,c));
			graph[b].add(new Edge(a,c));
		}
		int q = in.nextInt();
		for(int i = 0; i < q; ++i) {
			int u, v;
			u = in.nextInt()-1;
			v = in.nextInt()-1;
			solve(u,v);
		}
		in.close();
		
	}

	public static void solve(int u, int v) {
		if(graph[u] == null) {
			System.out.println(0);
		}
		else {
			int counter = 0;
			Set<Integer> checked = new HashSet<Integer>();
			for(Edge e: graph[u]) {
				visited = new boolean[graph.length];
				int color = e.color;
				if(checked.contains(color)) continue;
				Queue<Integer> q = new LinkedList<Integer>();
				q.add(u);
				visited[u] = true;
				
				while(!q.isEmpty()) {
					int curr = q.poll();
					if(curr == v) {
						counter++;
						break;
					}
					
					for(Edge edge: graph[curr]) {
						int c = edge.color;
						if(c != color) continue;
						if(!visited[edge.to]) {
							q.add(edge.to);
							visited[edge.to] = true;
						}
					}
				}
				checked.add(color);
			}
			System.out.println(counter);
		}
	}
	static class Edge {
		int to;
		int color;
		
		public Edge(int t, int c) {
			to = t; color = c;
		}
	}
}
