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>
#define N 100100
using namespace std;
int msk[N];
vector<int> g[N];
void DFS(int u,int p){
	for(int i=0;i<g[u].size();i++){
		int v=g[u][i];
		if(v!=p) DFS(v,u);
	}
	int d=1<<20;
	while(d){
		int c=0;
		for(int i=0;i<g[u].size();i++){
			int v=g[u][i];
			if(v!=p&&(msk[v]&d)) c++;
		}
		if(c>1){
			msk[u]+=d<<1;
			break;
		}
		else if(c==1) msk[u]|=d;
		d>>=1;
	}
	if(d==0) msk[u]++;
}
int main(){
	int n,a,b;
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		scanf("%d%d",&a,&b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	DFS(1,0);
	int ans=0;
	while((1<<ans+1)<=msk[1]) ans++;
	printf("%d\n",ans);
	return 0;
}