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

Mr. Kitayuta's Colorful Graph

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.

题意：有多条不同颜色的路径，求给定两点间相同颜色的路径条数。

思路：由于数据范围不大，可以开一个二维数组，使用二维并查集来处理多个相交的连通块。f[i][j]，i表示颜色，f[i][j]表示j的祖先。枚举每种color，当两点祖先相同时，即为有这种color路径使他们连通。

#include<stdio.h>
 
int f[][];
 
int find(int c,int x)
{
    return f[c][x]==x?x:f[c][x]=find(c,f[c][x]);
}
 
void join(int c,int x,int y)
{
    int fx=find(c,f[c][x]),fy=find(c,f[c][y]);
    if(fx!=fy) f[c][fy]=fx;
}
 
int main()
{
    int n,m,q,a,b,c,type,cc,i,j;
    scanf("%d%d",&n,&m);
    type=;
    for(i=;i<=;i++){
        for(j=;j<=;j++){
            f[i][j]=j;
        }
    }
    for(i=;i<=m;i++){
        scanf("%d%d%d",&a,&b,&c);
        if(c>type) type=c;
        join(c,a,b);
    }
    scanf("%d",&q);
    for(i=;i<=q;i++){
        scanf("%d%d",&a,&b);
        cc=;
        for(j=;j<=type;j++){
            if(find(j,f[j][a])==find(j,f[j][b])) cc++;
        }
        printf("%d\n",cc);
    }
    return ;
}