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[][];
 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=;i<=m;i++)
         for(int j=;j<=n;j++)
           fa[i][j]=j;
     int a,b,c;
     for(int i=;i<=m;i++)
     {
         scanf("%d %d %d",&a,&b,&c);
         unio(a,b,c);
     }
     int q;
     scanf("%d",&q);
     for(int i=;i<=q;i++)
     {
         scanf("%d %d",&a,&b);
         int ans=;
         for(int j=;j<=m;j++)
         {
             if(find(a,j)==find(b,j))
                 ans++;
         }
         printf("%d\n",ans);
     }
     return  ;
 }