Leaf Sets

You are given an undirected tree, consisting of $n$ vertices.

The vertex is called a leaf if it has exactly one vertex adjacent to it.

The distance between some pair of vertices is the number of edges in the shortest path between them.

Let’s call some set of leaves beautiful if the maximum distance between any pair of leaves in it is less or equal to $k$

You want to split all leaves into non-intersecting beautiful sets. What is the minimal number of sets in such a split?

Input

The first line contains two integers $n$ and $k (3≤n≤10^6, 1≤k≤10^6)$ — the number of vertices in the tree and the maximum distance between any pair of leaves in each beautiful set.

Each of the next $n−1$ lines contains two integers $v_i$ and $u_i (1≤v_i,u_i≤n)$ — the description of the $i$-th edge.

It is guaranteed that the given edges form a tree.

Output

Print a single integer — the minimal number of beautiful sets the split can have.

Examples

input

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

output

2

input

5 3 1 2 2 3 3 4 4 5

output

2

input

6 1 1 2 1 3 1 4 1 5 1 6

output

5

Note

Here is the graph for the first example:

題解

對於每個點維護最深的葉子離自己的距離，向上合併時，將所有兒子按最深的葉子離自己的距離排序，從深到淺遍歷嘗試合併，如果相鄰的兩個距離加起來大於$k$，便無法合併，向上傳也沒有意義，於是就將該兒子及其子樹單獨分成一個集合，答案+1，最後返回在刪去所有分離的兒子後最深的葉節點離自己的距離。

程式碼

#include<bits/stdc++.h>
#define add(a,b) mmp[a].push_back(b)
using namespace std;
const int M=1e6+5;
int ans,n,k;
vector<int>mmp[M];
int dfs(int f,int v)
{
	if(mmp[v].size()==1)return 0;
	vector<int>tmp;
	for(int i=mmp[v].size()-1;i>=0;--i)if(mmp[v][i]!=f)tmp.push_back(dfs(v,mmp[v][i])+1);
	sort(tmp.begin(),tmp.end());
	for(int siz;(siz=tmp.size())>1&&tmp[siz-1]+tmp[siz-2]>k;++ans,tmp.pop_back());
	return tmp.back();
}
void in(){scanf("%d%d",&n,&k);for(int i=1,a,b;i<n;++i)scanf("%d%d",&a,&b),add(a,b),add(b,a);}
void ac(){for(int i=1;i<=n;++i)if(mmp[i].size()>1)dfs(0,i),printf("%d",ans+1),exit(0);}
int main(){in(),ac();}