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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#define fo(i,a,b) for(int i=a;i<=b;++i)
#define fod(i,a,b) for(int i=a;i>=b;--i)
#define N 100005
using namespace std;
int nt[2*N],fs[N],dt[2*N],n,m,s[N],f[N][20],ans;
void link(int x,int y)
{
	nt[++m]=fs[x];
	dt[fs[x]=m]=y;
}
void dfs(int k,int fa)
{
	for(int i=fs[k];i;i=nt[i])
	{
		int p=dt[i];
		if(p!=fa)
		{
			dfs(p,k);
			fo(j,0,19) f[k][j]+=f[p][j];
		}
	}
	int lim=0;
	fod(j,19,0) if(f[k][j]>1)
	{
		lim=j+1;
		break;
	}
	while(f[k][lim]) lim++;
	s[k]=lim;
	fo(j,0,lim-1) f[k][j]=0;
	f[k][lim]++;
	ans=max(ans,s[k]);
}
int main()
{
	cin>>n;
	fo(i,1,n-1)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		link(x,y);
		link(y,x);
	}
	dfs(1,0);
	printf("%d",ans);
}