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<iostream>
#include<algorithm>
#include<vector>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef unordered_set<ll> set;
#define N 100010
#define D 20
ll n;
vector<ll> g[N];
set solve(ll x,ll from){
  vector<set> v;
  for(auto y:g[x])if(y!=from)v.push_back(solve(y,x));
  ll clash=-1;
  for(ll i=D;i-1;i--){
    int cnt=0;
    for(auto x:v)if(x.find(i)!=x.end())cnt++;
    if(cnt>=2){clash=i; break;}
  }
  ll gen;
  for(ll i=clash+1;;i++){
    int cnt=0;
    for(auto x:v)if(x.find(i)!=x.end())cnt++;
    if(cnt==0){gen=i; break;}
  }
  set res; res.insert(gen);
  for(ll i=gen+1;i<D;i++){
    int cnt=0;
    for(auto x:v)if(x.find(i)!=x.end())cnt++;
    if(cnt>0)res.insert(i);
  }
  return res;
}
int main(){
  cin>>n;
  for(int i=0;i<n-1;i++){
    ll a,b;cin>>a>>b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  set res=solve(1,0);
  ll ans=0;
  for(auto x:res)ans=max(ans,x);
  cout<<ans<<endl;
  return 0;
}
