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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;
#define SZ 666666
#define S 80
int fst[SZ],M=0,nxt[SZ],vb[SZ];
void ad_de(int a,int b)
{
	++M; nxt[M]=fst[a]; fst[a]=M; vb[M]=b;
}
void adde(int a,int b)
{ad_de(a,b);ad_de(b,a);}
bool f[100555][S];
int n,tmp[S];
void dfs(int x,int fa=0)
{
	for(int e=fst[x];e;e=nxt[e])
	{
		int b=vb[e]; if(b==fa) continue;
		dfs(b,x);
	}
	for(int i=0;i<S;++i) tmp[i]=0;
	for(int e=fst[x];e;e=nxt[e])
	{
		int b=vb[e]; if(b==fa) continue;
		for(int j=0;j<S;++j) tmp[j]+=f[b][j];
	}
	++tmp[0]; int r=-1;
	for(int j=0;j<S;++j)
		if(tmp[j]>=2)
			++tmp[j+1], r=j;
	for(int j=0;j<=r;++j) tmp[j]=0;
	for(int j=0;j<S;++j)
		f[x][j]=tmp[j];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;++i)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		adde(a,b);
	}
	dfs(1);
	for(int i=S-1;i>=0;--i)
	{
		if(f[1][i])
		{
			printf("%d\n",i);
			return 0;
		}
	}
}