CF286-B

B. Mr. Kitayuta's Colorful Graph

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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.

Sample test(s)

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.

给定n个点和m条边,点编号从1~n,每条边有:起点,终点和颜色

问给定任意两点,求连接这两点所有颜色的数量.(如:红色要连接两个点,那么这两点间必须存在一条颜色为红色的路径,即两点连通于红色)

求连通性很容易想到并查集.将每种颜色看成一个并查集,要求两点之间所有颜色时只需遍历所有颜色,看有多少种颜色能连通即可.

#include <iostream>
using namespace std;
class bingchaji
{
public:
    bingchaji()
    {
        init();
    }
    int father[200];
    void init()
    {
        for(int i=0;i<200;i++)
            father[i]=i;
    }
    void unit(int u,int v)
    {
        father[getfather(u)]=father[getfather(v)];
    }
    int getfather(int v)
    {
       if(father[v]==v)
           return v;
       return father[v]=getfather(father[v]);
    }
    bool same(int u,int v)
    {
        return getfather(u)==getfather(v);
    }
}color[200];

int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=m;i++)
    {
        int u,v,c;
        cin>>u>>v>>c;
        color[c].unit(u,v);
    }
    int k;
    cin>>k;
    for(int i=1;i<=k;i++)
    {
        int u,v;
        int res=0;
        cin>>u>>v;
        for(int j=1;j<=m;j++)
        {
            if(color[j].same(u,v))
                res++;
        }
        cout<<res<<endl;
    }
    return 0;
}