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>
const int N=2e5+10;
int ans,next[N],head[N],t[N],f[N],bz[N],tot; 
void ad(int x,int y)
{
	tot++;
	next[tot]=head[x];
	head[x]=tot;
	t[tot]=y;
}
int max(int x,int y)
{
	if (x>y) return x;
	return y;
}
int min(int x,int y)
{
	if (x<y) return x;
	return y;
}
void dfs(int x,int fa)
{
	int i=head[x],l=0;
	while (i)
	{
		if (t[i]==fa) 
		{
			i=next[i];
			continue;
		}
		dfs(t[i],x);
		l=l|(bz[x]&bz[t[i]]);
		bz[x]=bz[x]|bz[t[i]];
		i=next[i];
	} 
	if (!bz[x]) bz[x]=1;
	else 
	{
		while ((1<<f[x])<l||(1<<f[x])&bz[x]) f[x]++;
		bz[x]=(bz[x]>>f[x]<<f[x])|(1<<f[x]);
	}
	ans=max(ans,f[x]);
}
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		ad(x,y);
		ad(y,x);	
	}
	dfs(1,0); 
	printf("%d\n",ans); 
} 