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;

const int N=100100;
int n,ans;
vector<int>g[N];

int dfs(int x,int fa=-1){
	int ret=0,res=0;
	for(int i:g[x])if(i!=fa){
		int u=dfs(i,x);
		res|=ret&u;ret|=u;
	}
	int now=(res?32-__builtin_clz(res):0);
	now+=__builtin_ctz((ret>>now)+1);
	ans=max(ans,now);ret&=~((1<<now)-1);
	ret|=1<<now;return ret;
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<n;++i){
		int x,y;cin>>x>>y;
		g[x].push_back(y);g[y].push_back(x);
	}
	dfs(1);
	cout<<ans<<'\n';
	return 0;
}
