CodeForces - 505B Mr. Kitayuta's Colorful Graph（二维并查集。。）

Mr. Kitayuta has just bought an undirected graph consisting of n vertices and medges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color c_i, connecting vertex a_i and b_i.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers — u_i and v_i.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex u_i and vertex v_i 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 — a_i, b_i (1 ≤ a_i < b_i ≤ n) and c_i (1 ≤ c_i ≤ 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, (a_i, b_i, c_i) ≠ (a_j, b_j, c_j).

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 — u_i and v_i (1 ≤ u_i, v_i ≤ n). It is guaranteed that u_i ≠ v_i.

Output

For each query, print the answer in a separate line.

Example

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.

  原先我不敢相信，并查集还可以这么用，，仔细想想还真是的。。点还是点，只是多了对应边的的控制。

#include<stdio.h>
int n,m;
int f[200][200];
int getf(int v,int w)  //就是多了w，其他的都不变
{
    if(f[v][w]==v)
        return v;
    else
    {
        f[v][w]=getf(f[v][w],w);
        return f[v][w];
    }
}
void merge(int v,int u,int w)
{
    int t1=getf(v,w);
    int t2=getf(u,w);
    if(t1!=t2)
        f[t2][w]=t1;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
                f[i][j]=i;
        int a,b,c,q,x,y;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            merge(a,b,c);
        }
        scanf("%d",&q);
        for(int i=1; i<=q; i++)
        {
            int ans=0;
            scanf("%d%d",&x,&y);
            for(int j=1; j<=m; j++)
            {
                int a=getf(x,j);
                int b=getf(y,j);
                if(a==b)ans++;
            }
            printf("%d\n",ans);
        }
    }
}