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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=100005;

int n,cnt,last[N],bit[N],ans,bin[25];
struct edge{int to,next;}e[N*2];

int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

void addedge(int u,int v)
{
	e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;
	e[++cnt].to=u;e[cnt].next=last[v];last[v]=cnt;
}

void dfs(int x,int fa)
{
	int l=0;
	for (int i=last[x];i;i=e[i].next)
	{
		if (e[i].to==fa) continue;
		dfs(e[i].to,x);
		l|=bit[x]&bit[e[i].to];
		bit[x]|=bit[e[i].to];
	}
	int mx=20;
	while (mx>=0&&!(l&bin[mx])) mx--;
	mx++;
	while (bit[x]&bin[mx]) mx++;
	bit[x]|=bin[mx];bit[x]=bit[x]>>mx<<mx;
	ans=max(ans,mx);
}

int main()
{
	bin[0]=1;
	for (int i=1;i<=20;i++) bin[i]=bin[i-1]*2;
	n=read();
	for (int i=1;i<n;i++)
	{
		int x=read(),y=read();
		addedge(x,y);
	}
	dfs(1,0);
	printf("%d",ans);
	return 0;
}