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.

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.

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

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  ;

 }