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 <stdio.h>
#include <stdlib.h>
using namespace std;

int n,i,j,k,u,v,ans;
int son[100005],Next[200005],ed[200005],tot;
int S[100005];

void dfs(int x,int y)
{
	int i,j,k,A=0,B=0;
	for(i=son[x];i;i=Next[i])if(ed[i]!=y)dfs(ed[i],x);
	for(i=son[x];i;i=Next[i])
	if(ed[i]!=y)
	{
		B|=(A&S[ed[i]]);
		A|=S[ed[i]];
	}
	for(k=20;k>=0;--k)if(B&(1<<k))break;
	for(++k;(1<<k)&A;++k);
	if(k>ans)ans=k;
	S[x]=A|(1<<k);
	S[x]-=(S[x]&((1<<k)-1));
}

int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&u,&v);
		++tot;Next[tot]=son[u];son[u]=tot;ed[tot]=v;
		++tot;Next[tot]=son[v];son[v]=tot;ed[tot]=u;
	}
	dfs(1,0);
	printf("%d\n",ans);
}
