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;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
const int N=100005;
int n,tot,ans;
int h[N],ne[N<<1],to[N<<1],f[N][22];
void add(int x,int y) {to[++tot]=y,ne[tot]=h[x],h[x]=tot;}
void dfs(int x,int las) {
	for(RI i=h[x];i;i=ne[i]) {
		if(to[i]==las) continue;
		dfs(to[i],x);
		for(RI j=0;j<=20;++j) f[x][j]+=f[to[i]][j];
	}
	int p=0;
	for(RI i=20;i>=0;--i) if(f[x][i]>=2) {p=i+1;break;}
	while(f[x][p]) ++p;
	++f[x][p];for(RI i=0;i<p;++i) f[x][i]=0;
	ans=max(ans,p);
}
int main()
{
	int x,y;
	n=read();
	for(RI i=1;i<n;++i) x=read(),y=read(),add(x,y),add(y,x);
	dfs(1,0),printf("%d\n",ans);
	return 0;
}