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<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define fo(i,j,k) for(i=j;i<=k;i++)
#define fd(i,j,k) for(i=j;i>=k;i--)
#define cmax(a,b) (a=(a>b)?a:b)
#define cmin(a,b) (a=(a<b)?a:b)
typedef long long ll;
const int N=1e6+5,M=2e6+5,mo=1e9+7;
int n,ans,i,x,y;
int tt,nxt[N],b[N],fst[N];
void cr(int x,int y)
{
	tt++;
	b[tt]=y;
	nxt[tt]=fst[x];
	fst[x]=tt;
}
int f[N],sat[N];
void dfs(int x,int y)
{
	int lwb=0;
	for(int p=fst[x];p;p=nxt[p])
		if (b[p]!=y)
		{
			dfs(b[p],x);
			lwb|=sat[x]&sat[b[p]];
			sat[x]|=sat[b[p]];
		}
	f[x]=0;
	while ((1<<f[x])<=lwb||((1<<f[x])&sat[x])) f[x]++;
	sat[x]=(sat[x]>>f[x]<<f[x])|(1<<f[x]);
	cmax(ans,f[x]);
}
int main()
{
	scanf("%d",&n);
	fo(i,1,n-1)
	{
		scanf("%d %d",&x,&y);
		cr(x,y);
		cr(y,x);
	}
	dfs(1,0);
	printf("%d\n",ans);
}