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.*;
public class CodeforcesB505 {
	public static void main(String[] args){
		Scanner br = new Scanner(System.in);
		int n = br.nextInt();
		int m = br.nextInt();
		DisjointSet[] ds = new DisjointSet[m];
		for(int i = 0;i<m;i++){
			ds[i] = new DisjointSet(n);
		}
		
		for(int i = 0;i<m;i++){
			int s = br.nextInt()-1;
			int e = br.nextInt()-1;
			int c = br.nextInt()-1;
			ds[c].union(s, e);
		}
		
		int q = br.nextInt();
		for(int i = 0;i<q;i++){
			int s = br.nextInt()-1;
			int e = br.nextInt()-1;
			int ans = 0;
			for(int j = 0;j<m;j++){
				if(ds[j].find(s) == ds[j].find(e)){
					ans++;
				}
			}
			System.out.println(ans);
		}
	}
	public static class DisjointSet{
		int[] map;
		public DisjointSet(int n){
			map = new int[n+1];
			Arrays.fill(map, -1);
		}
		public int find(int x){
			if(map[x] < 0) return x;
			return map[x] = find(map[x]);
		}
		public void union(int a, int b){
			int roota = find(a);
			int rootb = find(b);
			if(roota == rootb) return;
			if(map[roota] < map[rootb]){
				map[roota] += map[rootb];
				map[rootb] = roota;
			}
			else{
				map[rootb]+=map[roota];
				map[roota] = rootb;
			}
		}
	}
}
