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 <cstdlib>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
#define efo(i,q) for(int i=A[q];i;i=B[i][0])
using namespace std;
const int N=100500;
int read(int &n)
{
	char ch=' ';int q=0,w=1;
	for(;(ch!='-')&&((ch<'0')||(ch>'9'));ch=getchar());
	if(ch=='-')w=-1,ch=getchar();
	for(;ch>='0' && ch<='9';ch=getchar())q=q*10+ch-48;n=q*w;return n;
}
int m,n,ans;
int B[2*N][2],A[N],B0;
int f[N];
int er[N];
void link(int q,int w)
{
	B[++B0][0]=A[q],A[q]=B0,B[B0][1]=w;
	B[++B0][0]=A[w],A[w]=B0,B[B0][1]=q;
}
void dfs(int q,int fa)
{
	int t=0;
	efo(i,q)if(B[i][1]!=fa)
	{
		dfs(B[i][1],q);
		t=max(t,f[q]&f[B[i][1]]);
		f[q]|=f[B[i][1]];
	}
	fo(i,1,22)if(er[i]>t&&!(f[q]&er[i])){t=i;break;}
	ans=max(ans,t);
	// printf("%d %d\n",q,t);
	f[q]=(f[q]&(er[23]-er[t]))|er[t];
}
int main()
{
	int q,w;
	er[1]=1;fo(i,2,23)er[i]=er[i-1]<<1;
	read(n);
	fo(i,1,n-1)read(q),read(w),link(q,w);
	dfs(1,0);
	printf("%d\n",ans-1);
	return 0;
}