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<algorithm>
using namespace std;
const int N=100005;
struct node { int y,next; } a[N<<1];
int f[N],s[N],head[N],i,j,k,n,m,an;
char c;

inline char getc()
{
	#define VV 10000000
	static char s[VV],*l=s,*r=s;
	if (l==r)
		l=s,r=s+fread(s,1,VV,stdin);
	return l==r?0:*l++;
}

int read(){ int z=0; do c=getc(); while (c<'0'||c>'9'); while (c>='0'&&c<='9') z*=10,z+=c-'0',c=getc(); return z; }

void jb(int x,int y)
{
	a[++k]=(node){y,head[x]},head[x]=k,
	a[++k]=(node){x,head[y]},head[y]=k;
}

void dfs(int w,int sp)
{
	int z=0;
	for (register int j=head[w];j;j=a[j].next)
		if ((j^1)!=sp)
		{
			int c=a[j].y;
			dfs(c,j),
			z|=s[w]&s[c],
			s[w]|=s[c];
		}
	if (s[w])
	{
		while ((1<<f[w])<=z||(s[w]>>f[w]&1)) ++f[w];
		s[w]>>=f[w],s[w]<<=f[w],s[w]|=1<<f[w];
	}
	else
		s[w]=1;
	an=max(an,f[w]);
}

int main()
{
	n=read(),k=1;
	for (i=1;i<n;++i) jb(read(),read());
	dfs(1,0);
	printf("%d\n",an);
	return 0;
}