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;
const int N=100005;
int cnt,head[N],n,u,v,Cnt[N][20],ans;
struct node{int to,next;}num[N<<1];
void add(int x,int y)
{
	num[++cnt].to=y;num[cnt].next=head[x];head[x]=cnt;
}
void dfs(int x,int fa)
{
	for (int i=head[x];i;i=num[i].next)
	  if (num[i].to!=fa) {
	     dfs(num[i].to,x);
	     for (int j=0;j<20;j++)
	       Cnt[x][j]+=Cnt[num[i].to][j];   
	  }
	int Min=0;
	for (int j=19;j>=0;j--) if (Cnt[x][j]>=2) {Min=j+1;break;}
	while (Cnt[x][Min]) Min++;
	Cnt[x][Min]=1;
	ans=max(ans,Min);
	for (int j=0;j<Min;j++) Cnt[x][j]=0;
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++) scanf("%d%d",&u,&v),add(u,v),add(v,u);
	dfs(1,-1);
	printf("%d\n",ans);
	return 0;
}