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 pb push_back

using namespace std;

const int N=100005;

vector<int>e[N];
int mask[N],ans,n,m,cnt[20];

void dfs(int x,int fa){
	for(int v:e[x])if(v!=fa)dfs(v,x);
	memset(cnt,0,sizeof(cnt));
	for(int v:e[x])if(v!=fa)for(int i=0;i<m;i++)cnt[i]+=mask[v]>>i&1;
	int p=-1,q=-1;
	for(int i=m-1;i>=0;i--){
		if(cnt[i]>1){p=i;break;}
		if(cnt[i]==0)q=i;
	}
	if(p==-1)p=q;
	mask[x]+=1<<p;
	for(int i=p;i<m;i++)if(cnt[i])mask[x]+=1<<i;
}

int main(){
	scanf("%d",&n);
	m=log(n)/log(2)+1;
	for(int i=1,u,v;i<n;i++)scanf("%d%d",&u,&v),e[u].pb(v),e[v].pb(u);
	dfs(1,0);
	while((1<<ans)<=mask[1])ans++;
	printf("%d\n",ans-1);
	return 0;
}