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<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

vector<int> edge[216000];
Int cnt[216000][20];
Int val[216000];
Int n, u, v;
void solve(Int x, Int last = -1){
  for(Int to:edge[x]){
    if(to == last)continue;
    solve(to, x);
    for(Int i = 0;i < 20;i++){
      cnt[x][i] += cnt[to][i];
    }
  }
  val[x] = 19;
  while(val[x] > 0 && cnt[x][val[x]] <= 1)val[x]--;
  while(cnt[x][val[x]] > 0)val[x]++;
  for(int i = 0;i < 20;i++){
    cnt[x][i] = !!cnt[x][i];
    if(i < val[x])cnt[x][i] = 0;
  }
  cnt[x][val[x]] = 1;
}

int main(){
  cin >> n;
  for(int i = 0;i < n-1;i++){
    cin >> u >> v;u--,v--;
    edge[u].push_back(v);
    edge[v].push_back(u);		   
  }
  solve(0);
  cout << *max_element(val, val + n) << endl;
  return 0;
}