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. 
//package com.pb.codeforces.practice;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class CF505B {
	
	public class UnionFind{
		
		int[] parent;
		int[] rank;
		public UnionFind(){
			parent = new int[101];
			rank = new int[101];
			makeSet();
		}
		
		public void makeSet() {
			for(int i=0; i<parent.length; i++)
				parent[i] = i;
		}
		
		public int find(int x) {
			int p = parent[x];
			if(x != p)
				parent[x] = find(parent[x]);
			return parent[x];
		}
		
		public void union(int x, int y) {
			int p1 = find(x);
			int p2 = find(y);
			if(p1 == p2) return;
			if(rank[p1] >= rank[p2]) {
				rank[p1] = rank[p1] == rank[p2] ? rank[p1] + 1 : rank[p1]; 
				parent[p2] = p1;
			}else {
				parent[p1] = p2;
			}
		}
	}
	
	public static void main(String[] args) {
		CF505B obj = new CF505B();
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		UnionFind uf[] = new UnionFind[101];
		HashSet<Integer>[] set = new HashSet[101];
		for(int i=0; i<101; i++) {
			uf[i] = obj.new UnionFind();
			set[i] = new HashSet<Integer>();
		}
		for(int i=0; i<m; i++) {
			int u = in.nextInt(); int v = in.nextInt(); int c = in.nextInt();
			uf[c].union(u, v);
			set[u].add(c);
			set[v].add(c);
		}
		int q = in.nextInt();
		for(int i=0; i<q; i++) {
			int u = in.nextInt(); int v = in.nextInt(); int cnt = 0;
			for(int c : set[u]) {
				if(uf[c].find(u) == uf[c].find(v))
					cnt++;
			}
			System.out.println(cnt);
		}
	}
}
