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<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;

typedef long long LL;
typedef double db;

int get(){
	char ch;
	while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');
	if (ch=='-'){
		int s=0;
		while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
		return -s;
	}
	int s=ch-'0';
	while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
	return s;
}

const int N = 1e+5+5;

int n;
struct edge{
	int x,nxt;
}e[N*2];
int h[N],tot;
int ans;
int f[N];

void inse(int x,int y){
	e[++tot].x=y;
	e[tot].nxt=h[x];
	h[x]=tot;
}

bool bz[N];

void dfs(int x){
	bz[x]=1;
	int pre=0,a=0;
	for(int p=h[x];p;p=e[p].nxt)
	if (!bz[e[p].x]){
		dfs(e[p].x);
		a|=(f[e[p].x]&pre);
		pre|=f[e[p].x];
	}
	fo(i,0,17)
	if ((1<<i)>a&&(pre&(1<<i))==0){
		pre=pre^(pre&((1<<i)-1));
		pre^=1<<i;
		ans=max(ans,i);
		break;
	}
	f[x]=pre;
}

int main(){
	n=get();
	fo(i,2,n){
		int x=get(),y=get();
		inse(x,y),inse(y,x);
	}
	dfs(1);
	printf("%d\n",ans);
	return 0;
}