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

struct node { int t,next; }a[200010];

int head[100010],bit[100010],f[100010],n,ans,tot;

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline void add(int x,int y) { a[++tot].t=y;a[tot].next=head[x];head[x]=tot; }

inline void dfs(int x,int y)
{
	bit[x]=f[x]=0;
	int hh=0;
	for (int i=head[x];i;i=a[i].next)
	{
		int t=a[i].t;
		if (t==y) continue;
		dfs(t,x);
		hh|=(bit[x]&bit[t]);
		bit[x]|=bit[t];
	}
	if (!bit[x]) bit[x]=1;
	else
	{
		while ((1<<f[x])<hh||((1<<f[x])&bit[x])) f[x]++;
		bit[x]=(bit[x]>>f[x]<<f[x])|(1<<f[x]);
	}
	ans=max(ans,f[x]);
}

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