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<bits/stdc++.h>
#define N 100005
using namespace std;
int n,fst[N],to[N*2],nxt[N*2],l,fa[N],b[N],Ans;
void link(int x,int y)
{
	to[++l]=y;nxt[l]=fst[x];fst[x]=l;
	to[++l]=x;nxt[l]=fst[y];fst[y]=l;
}
void dfs(int x)
{
	int tmp=0,s=0;
	for (int i=fst[x];i;i=nxt[i])
	if (to[i]!=fa[x])
	{
		fa[to[i]]=x;
		dfs(to[i]);
		s|=tmp&b[to[i]];
		tmp|=b[to[i]];
	}
	int t=0;
	while(((1<<t<s))||((1<<t)&tmp))t++;
	//cout<<t<<endl;
	b[x]=tmp^(1<<t)^(tmp&((1<<t)-1));
	Ans=max(Ans,t);
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		link(x,y);
	}
	dfs(1);
	/*int ans=0;
	while(Ans>1)Ans>>=1,ans++;*/
	printf("%d\n",Ans);
} 