Codeforces Round #286 (Div. 2) B 并查集

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.

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

qNote

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：n个点的图 m条边 两个点间可以有多条边 但是两点间相同颜色的边只能有一条  q个查询 判断 u-v间可以通过多少种颜色的边联通 

题解：并查集处理

#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <string>
#include <complex>
#define ll __int64
#define mod 1000000007
using namespace std;
int n,m;
int fa[105][105];
int find(int root,int c)
{
    if(fa[c][root]==root)
        return root;
    else
        return fa[c][root]=find(fa[c][root],c);
}
void unio(int a,int b,int c)
{
    int aa=find(a,c);
    int bb=find(b,c);
    if(aa!=bb)
        fa[c][aa]=bb;
}
int main()
{
    scanf("%d %d",&n,&m);
    for(int i=1;i<=m;i++)
        for(int j=1;j<=n;j++)
          fa[i][j]=j;
    int a,b,c;
    for(int i=1;i<=m;i++)
    {
        scanf("%d %d %d",&a,&b,&c);
        unio(a,b,c);
    }
    int q;
    scanf("%d",&q);
    for(int i=1;i<=q;i++)
    {
        scanf("%d %d",&a,&b);
        int ans=0;
        for(int j=1;j<=m;j++)
        {
            if(find(a,j)==find(b,j))
                ans++;
        }
        printf("%d
",ans);
    }
    return 0 ;
}