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;
vector<int> G[N];
int n,a[N];
void dfs(int t,int fa)
{
	int i,j,u=0,v=0;
	for(i=0;i<G[t].size();i++)
		if(G[t][i]!=fa){
			dfs(G[t][i],t);
			v|=u&a[G[t][i]];
			u|=a[G[t][i]];
		  }
	v=(v<<1)|1;
	for(i=30;i>=0;i--)
		if((v>>i)&1)
			for(j=i;j<=30;j++)
				if((~u>>j)&1){
					a[t]=(u^(u&((1<<j)-1)))|(1<<j);
					return ;
				  }
}
int main()
{
	int i,x,y;
	scanf("%d",&n);
	for(i=1;i<n;i++){
		scanf("%d %d",&x,&y);
		G[x].push_back(y);
		G[y].push_back(x);
	  }
	dfs(1,0);
	for(i=30;i>=0;i--)
		if((a[1]>>i)&1){
			printf("%d",i);
			return 0;
		  }
	return 0;
}
