Problem Description

Consider a un-rooted tree T which is not the biological significance of tree or plant, but a tree as an undirected graph in graph theory with n nodes, labelled from 1 to n. If you cannot understand the concept of a tree here, please omit this problem.
Now we decide to colour its nodes with k distinct colours, labelled from 1 to k. Then for each colour i = 1, 2, · · · , k, define Ei as the minimum subset of edges connecting all nodes coloured by i. If there is no node of the tree coloured by a specified colour i, Ei will be empty.
Try to decide a colour scheme to maximize the size of E1 ∩ E2 · · · ∩ Ek, and output its size.

 

Input

The first line of input contains an integer T (1 ≤ T ≤ 1000), indicating the total number of test cases.
For each case, the first line contains two positive integers n which is the size of the tree and k (k ≤ 500) which is the number of colours. Each of the following n - 1 lines contains two integers x and y describing an edge between them. We are sure that the given graph is a tree.
The summation of n in input is smaller than or equal to 200000.

 

Output

For each test case, output the maximum size of E1 ∩ E1 ... ∩ Ek.

 

Sample Input

 

Sample Output

 #include <stdio.h>

 #include <math.h>

 #include <string.h>

 #include <stdlib.h>

 #include <iostream>

 #include <sstream>

 #include <algorithm>

 #include <string>

 #include <queue>

 #include <ctime>

 #include <vector>

 using namespace std;

 const int maxn= 1e5+;

 const int maxm= 1e6+;

 const int inf = 0x3f3f3f3f;

 typedef long long ll;

 vector<int> G[maxn];

 int n,k,ans;

 int siz[maxn];

 void dfs(int x,int fa)

 {

     siz[x]=;

     for(int i=;i<G[x].size();i++)

     {

         int w=G[x][i];

         if(w!=fa)

         {

             dfs(w,x);

             siz[x]+=siz[w];

         }

     }

     if(siz[x]>=k&&n-siz[x]>=k)

         ans++;

 }

 int main()

 {

     int t;

     scanf("%d",&t);

     while(t--)

     {

         scanf("%d%d",&n,&k);

         int u,v;

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

             G[i].clear();

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

         {

             scanf("%d%d",&u,&v);

             G[u].push_back(v);

             G[v].push_back(u);

         }

         ans=;

         dfs(,);

         printf("%d\n",ans);

     }

 }