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 Test {
	static int n;
	static int m;
	static int count;
	static int[] fix;
	static ArrayList<Integer>[][] graph;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		m = sc.nextInt();
		graph = new ArrayList[m+1][n+1];
		fix = new int[n+1];
		for (int i = 0; i <= m; i++)
			for (int j = 0; j <= n; j++)
				graph[i][j] = new ArrayList<Integer>();
		for (int i = 1; i <= m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int c = sc.nextInt();
			graph[c][a].add(b);
			graph[c][b].add(a);
		}
		int q = sc.nextInt();
		for (int i = 1; i <= q; i++) {
			int q1 = sc.nextInt();
			int q2 = sc.nextInt();
			count = 0;
			for (int j = 1; j <= m; j++) {
				fix = new int[n+1];
				dfs(q1, q2, j);
			}
			System.out.println(count);
		}
	}
	public static void dfs(int a, int b, int c) {
		if (fix[a] == 1) return;
		if (a == b) {
			fix[a] = 1;
			count++;
			return;
		}
		fix[a] = 1;
		for (int i = 0; i < graph[c][a].size(); i++)
			dfs(graph[c][a].get(i), b, c);
	}
}