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>
#define N 200010
using namespace std;

inline 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;
}
struct edge
{
	int v,next;
}vs[N];
int n,mask[N][30],st[N],ee,Ans;
inline void addedge(int u,int v)
{
	vs[++ee].v=v;vs[ee].next=st[u];st[u]=ee;
}
inline void dfs(int rt,int pr)
{
	int cur=-1;
	for(int i=st[rt];i;i=vs[i].next)
	{
		if(vs[i].v==pr) continue;
		dfs(vs[i].v,rt);
		for(int j=0;j<30;j++)
		{
			if(mask[rt][j]&&mask[vs[i].v][j]) cur=max(cur,j);
			mask[rt][j]|=mask[vs[i].v][j];
		}
	}
	for(++cur;cur<30;++cur) if(!mask[rt][cur]) break;
	Ans=max(Ans,cur); mask[rt][cur]=1;
	for(--cur;cur>=0;cur--) mask[rt][cur]=0;
}
int main()
{
	n=read(); for(int i=1;i<n;i++)
	{
		int u=read(), v=read();
		addedge(u,v);
		addedge(v,u);
	}
	dfs(1,0); cout << Ans << endl;
}