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
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#include<bits/stdc++.h>
using namespace std;

const int MaxN=1e5+5;

int N,Ans;
int Cnt[MaxN][20];

vector<int> To[MaxN];

void DFS(int u,int fa){
	int i,k=0;
	for(auto v:To[u])
		if(v^fa){
			DFS(v,u);
			for(i=0;i<20;i++)
				Cnt[u][i]+=Cnt[v][i];
		}
	for(i=19;~i;i--)
		if(Cnt[u][i]>1){
			k=i+1;
			break;
		}
	for(;Cnt[u][k];k++);
	Ans=max(Ans,k);
	for(i=0;i<k;i++)
		Cnt[u][i]=0;
	Cnt[u][k]++;
}

int main(){
	int i,u,v;
	scanf("%d",&N);
	for(i=1;i<N;i++){
		scanf("%d%d",&u,&v);
		To[u].push_back(v);
		To[v].push_back(u);
	}
	DFS(1,0);
	printf("%d\n",Ans);
	return 0;
}