We will recursively define uninity of a tree, as follows: (Uni is a Japanese word for sea urchins.)

* A tree consisting of one vertex is a tree of uninity 0.
* Suppose there are zero or more trees of uninity k, and a vertex v. If a vertex is selected from each tree of uninity k and connected to v with an edge, the resulting tree is a tree of uninity k+1.



It can be shown that a tree of uninity k is also a tree of uninity k+1,k+2,..., and so forth.

You are given a tree consisting of N vertices. The vertices of the tree are numbered 1 through N, and the i-th of the N-1 edges connects vertices a_i and b_i.

Find the minimum k such that the given tree is a tree of uninity k.

Constraints

* 2 ≦ N ≦ 10^5
* 1 ≦ a_i, b_i ≦ N(1 ≦ i ≦ N-1)
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}


Output

Print the minimum k such that the given tree is a tree of uninity k.

Examples

Input

7
1 2
2 3
2 4
4 6
6 7
7 5


Output

2


Input

12
1 2
2 3
2 4
4 5
5 6
6 7
7 8
5 9
9 10
10 11
11 12


Output

3
#include<iostream>
#include<cstdio>
#include<cstring>
namespace nmy
{
	template<typename T>inline void check_max(T a,T &b){if(a>b)b=a;}
	template<typename T>inline void check_min(T a,T &b){if(a<b)b=a;}
	const int N=101000,M=N*2,K=20;
	int begin[N],next[M],to[M];
	int kjd[N],hdd[N];
	int n,e,ans;
	void add(int x,int y,bool k=1)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
		if(k)add(y,x,0);
	}
	void initialize()
	{
		scanf("%d",&n);
		for(int i=1,u,v;i<n;i++)
			scanf("%d%d",&u,&v),add(u,v);
	}
	void dfs(int p=1,int h=0)
	{
		int cnt[K+1];
		memset(cnt,0,sizeof(cnt));
		for(int i=begin[p],q;i;i=next[i])
			if((q=to[i])!=h)
			{
				dfs(q,p);
				for(int k=0;k<=K;k++)
					if(kjd[q]&(1<<k))cnt[k]++;

				for(int k=hdd[q];k<=K;k++)
					kjd[p]|=kjd[q]&(1<<k);
			}

		hdd[p]=K;
		for(int k=K;k>=0 && cnt[k]<=1;k--)
			if(!cnt[k])check_min(k,hdd[p]);
		kjd[p]|=1<<hdd[p];

		for(int k=0;k<hdd[p];k++)kjd[p]^=kjd[p]&(1<<k);

//		printf("f[%d] = %d\n",p,hdd[p]);

		check_max(hdd[p],ans);
	}
	void solve()
	{
		initialize();
		ans=0,dfs();
		printf("%d\n",ans);
	}
}
int main()
{
	nmy::solve();
	return 0;
}
