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>
#define N 100010
using namespace std;
int n,tote,to[N<<1],nxt[N<<1],con[N],w[N],ans;
void ins(int x,int y)
{
	to[++tote]=y;
	nxt[tote]=con[x];
	con[x]=tote;
}
void dfs(int v,int fa)
{
	int dep=-1;
	bool flag=0;
	for(int p=con[v];p;p=nxt[p])
		if(to[p]!=fa)
		{
			dfs(to[p],v);
			int tmp=w[v]&w[to[p]];
			for(int i=30;i>=0;i--)
				if((tmp>>i)&1) {dep=max(dep,i);break;}
			w[v]=w[v]|w[to[p]];
			flag=1;
		}	
	if(!flag) {w[v]=1;return ;}	
	for(int i=dep+1;i<=30;i++)
		if((w[v]>>i)&1) 
			dep=max(dep,i);
		else break;
	ans=max(ans,dep+1);
	int R=(1<<(dep+1)),T=((1<<30)-1)^(R-1);
	w[v]=(w[v]&T)|R;	
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		ins(x,y);ins(y,x);
	}
	dfs(1,0);
	printf("%d",ans);
	return 0;
}