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 {

	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		int n = cin.nextInt();
		int m = cin.nextInt();
		int maxc = -1;
		int[][][] dp = new int[101][101][101];
		for(int i = 0;i < m;i++) {
			int a,b,c;
			a = cin.nextInt();
			b = cin.nextInt();
			c = cin.nextInt();
			maxc = Math.max(c,maxc);
			dp[c][a][b] = dp[c][b][a] = 1;
		}
		
		//floyd
		for(int c = 1;c <= maxc;c++) //颜色
			for(int k = 1;k <= n;k++) //中间点
				for(int i = 1;i <= n;i++) //起始点
					for(int j = 1;j <= n;j++) //终止点
						//i和j之间没有颜色为c的连线 && i到k之间有颜色为c的连线 && k到j之间有颜色为c的连线
						if(dp[c][i][j] == 0 && dp[c][i][k] != 0 && dp[c][k][j] != 0)
							//则i到j之间就有颜色为c的连线
							dp[c][i][j] = dp[c][i][j] = 1;
		
		int q = cin.nextInt();
		while((q--) != 0) {
			int u = cin.nextInt();
			int v = cin.nextInt();
			int ans = 0;
			for(int i = 1;i <= maxc;i++)
				ans += dp[i][u][v];
			System.out.println(ans);
		}
	}
}