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 gc getchar()
#define ll long long
#define pb push_back
#define mk make_pair
#define rint register int
#define debug(x) cerr<<#x<<"="<<x<<"\n"
#define pa pair<int,int>
#define fir first
#define sec second
using namespace std;
inline int read(){int w=1,s=0;char ch=getchar();while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();}while(isdigit(ch)){s=s*10+ch-'0';ch=getchar();}return  w*s;}
struct node{int to,next;} e[290010];
int n,h[101010],tot,exist[101010][21],Ans;
inline void add(int from,int to){e[++tot].next=h[from];h[from]=tot;e[tot].to=to;}
inline void DFS(int now,int ffa){
	//debug(now);
	for(rint i=h[now];i;i=e[i].next){
		int to=e[i].to;if(to==ffa) continue;
		DFS(to,now);
		for(rint j=1;j<=20;++j) exist[now][j]+=exist[to][j];
	}
	int Min=0;
	for(rint i=1;i<=20;++i) if(exist[now][i]>=2) Min=i;
	//debug(Min);
	Min++;
	while(exist[now][Min]) Min++;
	Ans=max(Ans,Min);
	exist[now][Min]=1;
	for(rint i=1;i<=Min-1;++i) exist[now][i]=0;
}
int main()
{
	n=read();
	for(rint i=1;i<n;++i){
		int u=read(),v=read();
		add(u,v);add(v,u);
	}
	DFS(1,0);
	cout<<Ans-1<<"\n";
	return 0;
}
