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<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=beg[j];i;i=nex[i])
using namespace std;
int const mn=1e5+3;
int n,lg2,f[mn],gra,beg[mn],to[mn*2],nex[mn*2];
void ins(int u,int v){
	to[++gra]=v;
	nex[gra]=beg[u];
	beg[u]=gra;
}
void dfs(int p,int q){
	int tmp=0;
	fr(i,p)if(to[i]!=q){
		dfs(to[i],p);
		tmp|=f[p]&f[to[i]];
		f[p]|=f[to[i]];
	}
	int lim=0;
	fd(i,lg2,0)if(tmp&(1<<i)){lim=i+1;break;}
	fo(i,lim,lg2)if(!(f[p]&(1<<i)))
		{f[p]=((f[p]>>i)+1)<<i;return;}
}
int main(){
	scanf("%d",&n);
	fo(i,1,n-1){
		int u,v;
		scanf("%d%d",&u,&v);
		ins(u,v);
		ins(v,u);
	}
	lg2=log(n)/log(2)+1;
	dfs(1,0);
	fd(i,lg2,0)if(f[1]&(1<<i)){printf("%d",i);return 0;}
	return 0;
}
