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<cmath>
#include<algorithm>
#include<cstring>
using namespace std;

const int N=100005;

struct E{
	int to,next;
}mem[N<<1];
int num,n,x,y,ans;
int head[N],f[N][22];

void add(int x,int y){
	num++;
	mem[num].to=y; mem[num].next=head[x];
	head[x]=num;
}

void dfs(int k,int pre){
	int j,u,i,lim; lim=0;
	for (j=head[k];j;j=mem[j].next){
		u=mem[j].to;
		if (u==pre) continue;
		dfs(u,k);
		for (i=0;i<=20;i++) f[k][i]+=f[u][i];
	}
	for (i=0;i<=20;i++)
		if (f[k][i]>1) lim=i+1;
	while (f[k][lim]) lim++;
	for (i=0;i<lim;i++) f[k][i]=0;
	f[k][lim]++;
	ans=max(ans,lim);
}

int main(){
	int i;
	scanf("%d",&n);
	for (i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	dfs(1,0);
	printf("%d\n",ans);
	return 0;
}