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<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define N 101000
using namespace std;
int last[N],nxt[N*2],to[N*2],n,tot=1,f[N],g[N],ans=0;
void putin(int x,int y)
{
	nxt[++tot]=last[x];last[x]=tot;to[tot]=y;
}
void dg(int x,int fa)
{
	int jy=0;
	for(int i=last[x];i;i=nxt[i])
	{
		int y=to[i];if(y==fa) continue;
		dg(y,x);
		jy|=g[x]&g[y];
		g[x]=g[x]|g[y];
	}
	for(f[x]=0;((1<<f[x])<=jy)||((1<<f[x])&g[x]);f[x]++);
	ans=max(ans,f[x]);
	g[x]=g[x]>>f[x]<<f[x];
	g[x]|=1<<f[x];
}
int main()
{
	scanf("%d",&n);
	fo(i,1,n-1)
	{
		int x,y;scanf("%d%d",&x,&y);
		putin(x,y);putin(y,x);
	}
	dg(1,0);
	printf("%d\n",ans);
}