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<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
struct edges
{
	int to,next;
}edge[202020];
int first[101010],now;
void adde(int x,int y)
{
	now++;
	edge[now].to=y;
	edge[now].next=first[x];
	first[x]=now;
}
int s[101010],al=(1<<20)-1;
int pre[1<<20];
int lowbit(int x)
{
	return x&(-x);
}
int h=0;
void work(int p,int fa)
{
	int mk=1;
	s[p]=0;
	for(int e=first[p];e;e=edge[e].next)
		if(edge[e].to!=fa)
		{
			work(edge[e].to,p);
			mk=max(mk,pre[s[p]&s[edge[e].to]]);
			s[p]|=s[edge[e].to];
		}
	while(mk&s[p])
		mk<<=1;
	h=max(h,mk);
	s[p]&=(al^(mk-1));
	s[p]^=mk;
}
int main()
{
	int n,x,y,ans;
	scanf("%d",&n);
	pre[0]=0;
	pre[1]=1;
	for(int i=2;i<(1<<20);i++)
		pre[i]=pre[i>>1]<<1;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		adde(x,y);
		adde(y,x);
	}
	work(1,0);
	ans=-1;
	while(h)
	{
		ans++;
		h>>=1;
	}
	printf("%d\n",ans);
	return 0;
}