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;
using Int = long long;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  vector<vector<Int> > G(n);
  for(Int i=1;i<n;i++){
    Int a,b;
    cin>>a>>b;
    a--;b--;
    G[a].emplace_back(b);
    G[b].emplace_back(a);
  }
  
  function<Int(Int, Int)> dfs=[&](Int v,Int p){
    Int one=0,two=0;
    for(Int u:G[v]){
      if(u==p) continue;
      Int tmp=dfs(u,v);
      two|=(one&tmp);
      one|=tmp;
    }
    Int res=1;
    while(res<=two||(one&res)) res<<=1;
    return (one&~(res-1))|res;
  };
  
  Int res=dfs(0,-1);
  Int ans=31-__builtin_clz(res);
  cout<<ans<<endl;
  return 0;
}
