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>
using namespace std;
int n,f[110001][21],ans,s[110001];
vector<int> w[110001];
void dfs(int u,int p)
{
	for(int v:w[u])
		if(v!=p)
		{
			dfs(v,u);
			for(int j=0;j<18;j++)f[u][j]+=f[v][j];
		}
	int j=0;
	for(int i=17;i>=0;i--)
		if(f[u][i]>1){j=i+1;break;}
	while(f[u][j])j++;
	ans=max(ans,j);
	f[u][j]++;
	while(j--)f[u][j]=0;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		w[x].push_back(y);
		w[y].push_back(x);
	}
	dfs(1,-1);
	printf("%d\n",ans);
}