Mr. Kitayuta's Colorful Graph 【CodeForces - 505B 】

Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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.

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.

题意：

给出一个有向图，有不同的颜色的边，查询两点能够通过一种颜色连通的颜色的种类数。

代码：

#include<stdio.h>
#include<string.h>

int n,m;
int x,y,z,a;
int s[110][110][110];

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++)
	{
		scanf("%d%d%d",&x,&y,&z);
		s[x][y][z]=1;             //标记
		s[y][x][z]=1;
	}
	
	for(int i=1;i<=m;i++)
	   for(int j=1;j<=n;j++)
	      for(int k=1;k<=n;k++)
	          for(int w=1;w<=n;w++)
	          {
	          	if(s[k][j][i] && s[j][w][i])   //进入循环的条件
	          	{
	          		s[k][w][i]=1;
	          		s[w][k][i]=1;
				}
			  }
			  
		scanf("%d",&a);
		while(a--)
		{
			int ans=0;
			scanf("%d%d",&x,&y);     
			for(int i=1;i<=m;i++)   //循环
			   ans=ans+s[x][y][i];
			printf("%d\n",ans);
		}	 
	return 0; 
}