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 <math.h>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <stack>
#include <string>
#include <queue>
#include <map>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <sstream>
#include <set>
using namespace std;

int finder(const vector<vector<int> >& G, int now, int prv) {
  vector<int> bits;
  for (auto nxt : G[now]) {
    if (nxt == prv) continue;
    bits.push_back(finder(G, nxt, now));
  }
  int res = 0;
  for (int bit=20; bit>=0; bit--) {
    int cnt = 0;
    for (auto p : bits) {
      if (p&(1<<bit)) cnt ++;
    }
    if (cnt == 1) res += 1<<bit;
    else if (cnt > 1) {
      res += 1<<(bit+1);
      return res;
    }
  }
  res ++;
  return res;
}

int main()
{
  int N;
  cin >> N;
  vector<vector<int> > G(N);
  for (int i=0; i<N-1; i++) {
    int a, b;
    cin >> a >> b;
    G[a-1].push_back(b-1);
    G[b-1].push_back(a-1);
  }

  int res = finder(G, 0, -1);
  int ret = 0;
  while (res > 1) {
    res /= 2;
    ret ++;
  }
  printf("%d\n", ret);
}
