The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The cows didn't have time to rebuild any extra roads, so now there is exactly one way to get from any given barn to any other barn. Thus, the farm transportation system can be represented as a tree.

Farmer John wants to know how much damage another earthquake could do. He wants to know the minimum number of roads whose destruction would isolate a subtree of exactly P (1 <= P <= N) barns from the rest of the barns.

Input

* Line 1: Two integers, N and P

* Lines 2..N: N-1 lines, each with two integers I and J. Node I is node J's parent in the tree of roads.

Output

A single line containing the integer that is the minimum number of roads that need to be destroyed for a subtree of P nodes to be isolated.

Sample Input

11 6
1 2
1 3
1 4
1 5
2 6
2 7
2 8
4 9
4 10
4 11

Sample Output

2

Hint

[A subtree with nodes (1, 2, 3, 6, 7, 8) will become isolated if roads 1-4 and 1-5 are destroyed.]

題解：dp[i][j] 表示以i這個節點 和子代連線 如果儲存j個節點的話 需要去除多少遍  顯然dp[i][1] 就是i子代的個數 因為你只能從一個子代那裡 更新一個狀態  一次  顯然這是一個分組揹包  因為該節點必須要所以更新到2就可以了 在最後取最小值時，若該點不是根節點 需要在結果的基礎上+1，因為該點還與父節點連線著

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
const int N=150+10;
vector<int> v[N];
int dp[N][N],in[N];
int n,m;
void dfs(int u)
{
	dp[u][1]=v[u].size();//
	for(int i=0;i<v[u].size();i++)
	{
		int to=v[u][i];
		dfs(to);
		for(int j=m;j>=2;j--)
			for(int k=1;k<j;k++)
				dp[u][j]=min(dp[u][j],dp[to][k]+dp[u][j-k]-1);// 這個地方 坑點 若和子代連線 那麼這條邊就不去了 
	}	
} 
int main()
{
	int x,y;
	while(~scanf("%d%d",&n,&m))
	{
		for(int i=1;i<=n;i++) v[i].clear();
		for(int i=1;i<n;i++)
		{
			scanf("%d%d",&x,&y);
			v[x].push_back(y);
		}
		memset(dp,INF,sizeof(dp));
		dfs(1);
		int ans=dp[1][m];
		for(int i=2;i<=n;i++) ans=min(ans,dp[i][m]+1);
		printf("%d\n",ans);
	}
	return 0;
}