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;
 
struct edge{int s,t,n;}e[200010];
int n,h[200010],f[200010];
 
void dfs(int x,int fa)
{
	int s=0,sum=0,k=1,p=0;
	for (int i=h[x],y; y=e[i].t,i; i=e[i].n)
		if (y!=fa)  dfs(y,x),s|=sum&f[y],sum|=f[y];
	while ((k<s)||(sum&k))  {k<<=1;p++;}
	//f[x]=(sum^(sum&(k-1)))|k;
f[x]=(sum>>(p)<<(p))|k;
}
 
void work()
{
	scanf("%d",&n);
	for (int i=1,u,v,tot=0; i<n; i++)
		{
			scanf("%d %d",&u,&v);
			e[++tot]=(edge){u,v,h[u]},h[u]=tot;
			e[++tot]=(edge){v,u,h[v]},h[v]=tot;
		}
	dfs(1,0);
	printf("%d",int(log2(f[1])));
}
 
int main()
{
	work();
	return 0;
}