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.

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <cmath>

 #include <map>

 #include <set>

 #include <stack>

 #include <queue>

 #include <string>

 #include <vector>

 using namespace std;

 const double EXP=1e-;

 const double PI=acos(-1.0);

 const int INF=0x7fffffff;

 const int MS=;

 int fa[MS][MS];

 void init()

 {

     for(int i=;i<MS;i++)

         for(int j=;j<MS;j++)

             fa[i][j]=j;

 }

 int find(int c,int a)

 {

     if(fa[c][a]==a)

         return a;

     return fa[c][a]=find(c,fa[c][a]);

 }

 void make_union(int c,int a,int b)

 {

     a=find(c,a);

     b=find(c,b);

     if(a==b)

         return ;

     fa[c][b]=fa[c][a];

 }

 int main()

 {

     int n,m,q,a,b,c;

     init();

     cin>>n>>m;

     for(int i=;i<m;i++)

     {

         cin>>a>>b>>c;

         make_union(c,a,b);

     }

     cin>>q;

     while(q--)

     {

         int ans=;

         cin>>a>>b;

         for(c=;c<=m;c++)

             if(find(c,a)==find(c,b))

                 ans++;

         cout<<ans<<endl;

     }

     return ;

 }