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<bits/stdc++.h>
using namespace std;
int n,f[100010][25];
vector<int> a[100010];
inline void dp(int i,int l)
{
    int j,k;
    for(j=0;j<a[i].size();j++)
      if(a[i][j]!=l)
        {
         dp(a[i][j],i);
         for(k=1;k<=20;k++)
           f[i][k]+=f[a[i][j]][k];
        }
    for(j=20;j>0;j--)
      if(f[i][j]>1)
        break;
    for(j++;f[i][j];j++);
    f[i][j]=1;
    for(j--;j>0;j--)
      f[i][j]=0;
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k;
	scanf("%d",&n);
	for(i=1;i<n;i++)
	  {
       scanf("%d%d",&j,&k);
       a[j].push_back(k);
       a[k].push_back(j);
      }
    dp(1,0);
    for(i=20;!f[1][i];i--);
    printf("%d\n",i-1);
	return 0;
}
