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;
#define N 1111116
#define next nxt

int n,x[N],y[N],next[N],p[N],f[N],g[N],ans;

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-')  f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

void travel(int x){
	int tmp=0,t=0;
	for (int t=p[x];t;t=next[t])
		if (f[x]!=y[t]){
			f[y[t]]=x;travel(y[t]);
			tmp|=g[x]&g[y[t]];g[x]|=g[y[t]];
		}
	while ((1<<t)<=tmp||(1<<t)&g[x]) t++;
	g[x]=(g[x]>>t<<t)|(1<<t);
	ans=max(ans,t);
}

int main(){
	n=read();
	for (int i=1;i<=n-1;i++){
		x[2*i-1]=y[2*i]=read();
		y[2*i-1]=x[2*i]=read();
		next[2*i-1]=p[x[2*i-1]];p[x[2*i-1]]=2*i-1;
		next[2*i]=p[x[2*i]];p[x[2*i]]=2*i;
	}
	travel(1);printf("%d\n",ans);
	return 0;
}