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<vector>
#include<cstdlib>
using namespace std;
int n;
vector<int>G[100010];
int state[100010];
int ans;
void dfs(int x,int p)
{
	int all[22];
	for(int i=0;i<=20;i++)all[i]=0;
	state[x]=0;
	for(int i=0;i<G[x].size();i++)
	{
		int y=G[x][i];
		if(y==p)continue;
		dfs(y,x);
		state[x]|=state[y];
	    for(int j=0;j<=20;j++)
	    {
	    	if(state[y]>>j&1)
	    	{
	    		all[j]++;
	    	}
	    }
	}
	int res=0;
	for(int i=20;i>=0;i--)
	{
		if(all[i]>1)
		{
			res=i+1;
			break;
		}
	}
	for(int i=res;i<=20;i++)
	{
		if(state[x]>>i&1)continue;
		res=i;
		break;
	}
	state[x]|=(1<<res);
	for(int i=0;i<res;i++)if(state[x]>>i&1)state[x]^=(1<<i);
	ans=max(ans,res);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		G[a].push_back(b);
		G[b].push_back(a); 
	}
	dfs(1,0);
	printf("%d",ans);
	return 0;
}