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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
const int N=2e5;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
int head[N],nxt[N],to[N];
int f[N];
//给每个点一个标号，使得任意两个相同标号的点的路径上存在点标号大于这两个点
//每次贪心选择最小的标号
inline void dfs(int k,int fa) {
	int s=0,all=0;
	for (int i=head[k];i;i=nxt[i])
		if (to[i]!=fa) {
			dfs(to[i],k);
			all|=s&f[to[i]];
			s|=f[to[i]];
		}
	(all<<=1)|=1;
	for (int i=30;i>=0;i--)
		if (all>>i&1) {
			while (s>>i&1) i++;
			f[k]=(s^(s&((1<<i)-1)))|(1<<i);
			return;
		}
}
int main()
{
	int n=gi(),i,a,b,tot=0;
	for (i=1;i<n;i++) {
		a=gi(),b=gi();
		to[++tot]=b,nxt[tot]=head[a],head[a]=tot;
		to[++tot]=a,nxt[tot]=head[b],head[b]=tot;
	}
	dfs(1,0);
	for (i=30;i>=0;i--)
		if (f[1]>>i&1)
			return printf("%d\n",i),0;
	return 0;
}
