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 dp[100001][20];

vector<int> e[100001];


void dfs(int x,int p){
	for(int it:e[x]){
		if(it!=p){
			dfs(it,x);
			for(int i=0;i<=19;i++)
				dp[x][i]+=dp[it][i];
		}
	}
	int who;
	for(who=19;who>=0;who--)
		if(dp[x][who]>1)
			break;
	who++;
	while(dp[x][who])	who++;
	dp[x][who]++;
	for(int i=0;i<who;i++)
		dp[x][i]=0;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	for(int i=1,u,v;i<n;i++){
		cin>>u>>v;
		e[u].emplace_back(v);
		e[v].emplace_back(u);
	}
	dfs(1,0);
	for(int i=19;;i--)
		if(dp[1][i]){
			cout<<i<<endl;
			return 0;
		}
}
