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<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

inline int read()
{
	int x=0;char ch=getchar();
	while(ch<'0' || '9'<ch)ch=getchar();
	while('0'<=ch && ch<='9')x=x*10+(ch^48),ch=getchar();
	return x;
}

const int N=1e5+9;

unsigned int bit[N];
int n,f[N],ans;
int to[N<<1],nxt[N<<1],beg[N],tot;

inline void add(int u,int v)
{
	to[++tot]=v;
	nxt[tot]=beg[u];
	beg[u]=tot;
}

inline void dfs(int u,int fa)
{
	bit[u]=f[u]=0;
	int las=0;
	for(int i=beg[u],v;i;i=nxt[i])
		if((v=to[i])!=fa)
		{
			dfs(v,u);
			las|=bit[v]&bit[u];
			bit[u]|=bit[v];
		}
	if(bit[u]==0)
		bit[u]=1;
	else
	{
		while((1<<f[u])<las || bit[u]&(1<<f[u]))f[u]++;
		bit[u]=(bit[u]>>f[u]<<f[u])|(1<<f[u]);
	}
	if(ans<f[u])ans=f[u];
}

int main()
{
	n=read();
	for(int i=1,u,v;i<n;i++)
	{
		u=read();v=read();
		add(u,v);add(v,u);
	}

	dfs(1,0);
	printf("%d\n",ans);
	return 0;
}
