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[505000], g[22];
vector<int>G[505000];

int mg(int x,int y){
  for (int t;t=(x&y);){
    x+=t&-t;
  }
  return min(10001000,x|y);
}

void dfs(int x,int fff){
  for (auto y:G[x])
    if (y!=fff)
      dfs(y,x);
  memset(g,0,sizeof g);
  for (auto y:G[x])
    if (y!=fff){
      for (int i=0;i<=20;++i){
	int tmp=f[y]>>i<<i;
	if (tmp>>i&1) tmp+=1<<i;
	g[i]=mg(g[i],tmp);
      }
    }
  f[x]=998244353;
  for (int i=0;i<=20;++i)
    f[x]=min(f[x],g[i]|1<<i);
}

int main(){
  cin>>n; int x, y;
  for (int tim=n-1;tim--;){
    cin>>x>>y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  dfs(1,0);
  int ans=0;
  for (int t=f[1];t;t>>=1) ++ans;
  cout<<ans-1<<endl;
}
