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>
#define N (100000+1000)
using namespace std;

struct Edge{int to,next;}edge[N<<1];
int n,u,v,dep[N],f[N][22],ans;
int head[N],num_edge;

void add(int u,int v)
{
	edge[++num_edge].to=v;
	edge[num_edge].next=head[u];
	head[u]=num_edge;
}

void Dfs(int x,int fa)
{
	for (int i=head[x]; i; i=edge[i].next)
		if (edge[i].to!=fa)
		{
			Dfs(edge[i].to,x);
			for (int j=0; j<=20; ++j)
				f[x][j]+=f[edge[i].to][j];
		}
	int lim=0;
	for (int i=20; i>=0; --i)
		if (f[x][i]>1) {lim=i+1; break;}
	while (f[x][lim]) lim++;
	dep[x]=lim;
	for (int j=0; j<=lim-1; ++j) f[x][j]=0;
	f[x][lim]++; ans=max(ans,dep[x]);
}

int main()
{
	scanf("%d",&n);
	for (int i=1; i<=n-1; ++i)
	{
		scanf("%d%d",&u,&v);
		add(u,v); add(v,u);
	}
	Dfs(1,0);
	printf("%d\n",ans);
}