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;
typedef vector<int> vi;
typedef vi::iterator vit;
typedef vi::reverse_iterator vrit;
#define pb push_back
const int MX=100005;
vi G[MX];
int n;
int dp(int k,int f){
	int v=0,v2=0;
	for(auto i:G[k])if(i!=f){
		int r=dp(i,k);
		v2|=(r&v),v|=r;
	}
	int it=0;while((1<<it)<=v2)it++;
	while((v>>it)&1)it++;
	return (v|(1<<it))&(~((1<<it)-1));
}
int main(){
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<n;i++){
		int a,b;cin>>a>>b;
		G[a].pb(b),G[b].pb(a);
	}
	cout<<31-__builtin_clz(dp(1,0))<<endl;
	return 0;
}