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.Scanner;


public class Main_Round286Div2_B {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int m = sc.nextInt();

		UnionFind[] uf = new UnionFind[m];
		for (int i = 0; i < m; i++) {
			uf[i] = new UnionFind(n);
		}
		for (int i = 0; i < m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int c = sc.nextInt();
			
			uf[c - 1].union(a - 1, b - 1);
		}

		int q = sc.nextInt();
		for (int i = 0; i < q; i++) {
			int ret = 0;
			int u = sc.nextInt();
			int v = sc.nextInt();
			for (int j = 0; j < m; j++) {
				if (uf[j].same(u - 1, v - 1)) {
					ret++;
				}
			}
			
			System.out.println(ret);
		}

		sc.close();
	}

	private static class UnionFind {
		int[] parent;

		UnionFind(int n) {
			parent = new int[n];
			for (int i = 0; i < n; i++) {
				parent[i] = i;
			}
		}
		
		int find(int x) {
			if (parent[x] == x) {
				return x;
			} else {
				return parent[x] = find(parent[x]);
			}
		}
		
		boolean same(int x, int y) {
			return find(x) == find(y);
		}
		
		void union(int x, int y) {
			x = find(x);
			y = find(y);
			if (x != y) {
				parent[x] = y;
			}
			
			return;
		}

		// 異なる集合の数
		@SuppressWarnings("unused")
		int count() {
			int ret = 0;
			for (int i = 0; i < parent.length; i++) {
				if (find(i) == i) {
					ret++;
				}
			}
			
			return ret;
		}
	}
}
