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 <cstring>
#include <algorithm>

using namespace std;

const int N=524444;

typedef long long LL;

int n,f[N],ans,g[N],h[N],e[N],nxt[N],tot;

char c;

int read()
{
    int x=0,sig=1;
    for (c=getchar();c<'0' || c>'9';c=getchar()) if (c=='-') sig=-1;
    for (;c>='0' && c<='9';c=getchar()) x=x*10+c-48;
    return x*sig;
}

void Add(int x,int y)
{
	e[++tot]=y; nxt[tot]=h[x]; h[x]=tot;
}

void dfs(int x,int y)
{
	int i,j,k;
	for (i=h[x],j=k=0;i;i=nxt[i]) if (e[i]!=y)
	{
		dfs(e[i],x);
		if ((j&f[e[i]])>0) k=max(k,g[j&f[e[i]]]+1);
		j|=f[e[i]];
	}
	//for (i=0;((1<<i)&j)>0;i++);
	//k=max(k,i);
	for (;((1<<k)&j)>0;k++);
	ans=max(ans,k);
	f[x]=1<<k;
	for (;g[j]>k;j^=(1<<g[j])) f[x]|=(1<<g[j]);
}

int main()
{
	n=read();
	for (int i=1,x,y;i<n;i++)
	{
		x=read(); y=read();
		Add(x,y); Add(y,x);
	}
	for (int i=1,j=0;i<N;i<<=1,j++) g[i]=j;
	for (int i=3;i<N;i++) if (!g[i]) g[i]=g[i-1];
	dfs(1,0);
	printf("%d\n",ans);
    return 0;
}