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<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#define MAXN 100000
using namespace std;
typedef long long LL;
vector<int> G[MAXN+5];
int N,ans;
int cnt[MAXN+5][21];
void DFS(int u,int fa)
{
	for(int i=0;i<(int)G[u].size();i++)
	{
		int v=G[u][i];
		if(v==fa)
			continue;
		DFS(v,u);
		for(int j=0;j<20;j++)
			cnt[u][j]+=cnt[v][j];
	}
	int val=0;
	for(int i=20;i>=0;i--)
		if(cnt[u][i]>=2)
		{
			val=i+1;
			break;
		}
	while(cnt[u][val]==1)
		val++;
	ans=max(ans,val);
	cnt[u][val]++;
	for(int i=0;i<val;i++)
		cnt[u][i]=0;
}
int main()
{
//	freopen("uninity.in","r",stdin);
//	freopen("uninity.out","w",stdout);
	scanf("%d",&N);
	int u,v;
	for(int i=1;i<N;i++)
	{
		scanf("%d %d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	DFS(1,-1);
	printf("%d\n",ans);
	return 0;
}