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.Scanner;

public class CF286B {

	static boolean mx[][][];
	static int n;
	static int m;
	static int a;
	static int b;
	static int count;

	static boolean mark[];

	static void justBFS(int v, int c) {
		int[] queue = new int[n]; // �������
		int qH = 0; // ������ �������
		int qT = 0; // �����

		/* <��������� ������� v> */
		mark[v] = true; // �������� �������� �������
		queue[qT++] = v; // �������� �� � �������

		while (qH < qT) { // ���� ������� �� �����
			v = queue[qH++]; // ��������� ������� �������
			for (int nv = 0; nv < n; nv++) { // ���������� �������
				if (!mark[nv] && mx[c][v][nv]) { 
					if (nv==b){
						count++;
						return;
					}
				/* <��������� ������� nv> */
					mark[nv] = true; // �������� ��
					queue[qT++] = nv; // � ��������� � �������
				}
			}
		}
	}

	static void justDFS(int v, int c) {
		mark[v] = true;
		for (int nv = 0; nv < n; nv++) {
			if (!mark[nv] && mx[c][v][nv]) {
				if (nv == a) {
					count++;
					return;
				}
				justDFS(nv, c);
			}
		}
	}

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		n = scan.nextInt();
		m = scan.nextInt();
		mx = new boolean[m][n][n];

		for (int i = 0; i < m; i++) {
			int uu = scan.nextInt() - 1;
			int vv = scan.nextInt() - 1;
			int cc = scan.nextInt() - 1;
			mx[cc][uu][vv] = true;
			mx[cc][vv][uu] = true;
		}
		int q = scan.nextInt();
		for (int ii = 0; ii < q; ii++) {
			a = scan.nextInt() - 1;
			b = scan.nextInt() - 1;
			count = 0;
			for (int i = 0; i < m; i++) {
				mark = new boolean[n];
				justBFS(a, i);
			}
			System.out.println(count);

		}

	}

}
