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 <bits/stdc++.h>
#define N 100005
using namespace std;
struct edge{
	int k,next;
}e[N<<1];
int n,home[N],cnt=-1,lg2[N<<1];
void add(int x,int y){
	cnt++;
	e[cnt].k=y;
	e[cnt].next=home[x];
	home[x]=cnt;
}
int f[N],sum[25];
void dfs(int k,int F){
	for(int i=home[k];~i;i=e[i].next)if(e[i].k!=F){
		dfs(e[i].k,k);
	}
	memset(sum,0,sizeof(sum));
	for(int i=home[k];~i;i=e[i].next)if(e[i].k!=F){
		int g=f[e[i].k],t;
		while(g) t=g&-g,sum[lg2[t]]++,g-=t;
	}
	int o=0;
	for(int i=20;~i;i--){
		if(sum[i]){
			if(!o) o=(1<<i);
			if(sum[i]>1){
				i++;
				while(sum[i]){
					f[k]^=(1<<i);
					i++;
				}
				f[k]|=(1<<i);
				return;
			}
			f[k]|=(1<<i);
		}else if(o){
			int h=0,g;
			while(i>=0){
				if(!sum[i]) g=(1<<i)|h;
				else h|=(1<<i);
				if(sum[i]>1) break;
				i--;
			}
			f[k]|=g;
			return;
		}
	}
	f[k]=o+o;
	if(!f[k]) f[k]=1;
}
int main(){
	//freopen("test.in","r",stdin);
	memset(home,-1,sizeof(home));
	for(int i=2;i<=200000;i++) lg2[i]=lg2[i>>1]+1;
	scanf("%d",&n);
	for(int i=1,x,y;i<n;i++){
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	dfs(1,0);
	printf("%d\n",lg2[f[1]]);
	return 0;
}