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<vector>
#include<algorithm>
using namespace std;
const int MAXN=100005;

int N;
vector<int> adj[MAXN];
int val[MAXN],ans;

void dfs(int u,int fa=0)
{
	int mk=0;
	for(int i=0;i<(int)adj[u].size();i++)
	{
		int v=adj[u][i];
		if(v==fa)
			continue;
		dfs(v,u);
		mk|=(val[u]&val[v]);
		val[u]|=val[v];
	}
	if(val[u]==0)
		val[u]=1;
	else
	{
		int id=0;
		while((1<<id)<mk||(val[u]&(1<<id)))
			id++;
		val[u]=((val[u]>>id)<<id)|(1<<id);
		ans=max(ans,id);
	}
	//printf("val[%d]=%d\n",u,val[u]);
}

int main()
{
	scanf("%d",&N);
	for(int i=1;i<N;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		adj[a].push_back(b);
		adj[b].push_back(a);
	}
	dfs(1);
	printf("%d\n",ans);
	
	return 0;
}