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>
#define N 100010
using namespace std;
int nxt[N<<1],to[N<<1],head[N],cnt;
void add(int u,int v)
{
	nxt[++cnt]=head[u];
	to[cnt]=v;
	head[u]=cnt;
}
int f[N][21],ans;
void dfs(int u,int pre)
{
	for(int i=head[u];i;i=nxt[i])
	{
		int v=to[i];
		if(v==pre) continue;
		dfs(v,u);
		for(int j=0;j<=20;j++) f[u][j]+=f[v][j];
	}
	int pmin=0;
	for(int j=20;j>=0;j--)
	if(pmin)f[u][j]=0;
	else if(f[u][j]>=2) pmin=j;
	while(!pmin || f[u][pmin]>=2) f[u][pmin]=0,f[u][++pmin]++;
	ans=max(ans,pmin);
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	dfs(1,0);
	printf("%d",ans-1);
	return 0;
}
