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 fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
const int maxn=1e5+5;
inline int read(){
	int n=0,f=1;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if (c=='-') c=getchar(),f=-1;
	for(;c>='0'&&c<='9';c=getchar()) n=n*10+c-48;
	return n*f;
}
int i,j,n,k[maxn],ne[maxn<<1],g[maxn<<1],num,f[maxn];
void add(int x,int y){
	ne[++num]=k[x],k[x]=num,g[num]=y; 
}
void dfs(int x,int y){
	int i,s=0,now=0;
	for(i=k[x];i;i=ne[i]){
		int go=g[i];
		if (go!=y){
			dfs(go,x);
			now=max(now,(s&f[go]));
			s|=f[go];
		}
	}
	if (!now) {
		fo(i,0,20) if (!(s&(1<<i))) break;
		f[x]=(1<<i)|((s>>i)<<i); 
	}else{
		fo(i,0,20) if (!(s&(1<<i))&&((1<<i)>now)) break;
		f[x]=(1<<i)|((s>>i)<<i);
	} 
}
int main(){
	n=read();
	fo(i,1,n-1){
		int x=read(),y=read();
		add(x,y),add(y,x);
	}
	dfs(1,0);
	fd(i,20,0) if (f[1]&(1<<i)) break;
	printf("%d\n",i);
}