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
/**
 *    author:  tourist
 *    created: 20.11.2019 16:54:11       
**/
#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    --x; --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  vector<int> mask(n);
  int ans = 0;
  function<void(int, int)> Dfs = [&](int v, int pr) {
    int lowest = 0;
    for (int u : g[v]) {
      if (u == pr) {
        continue;
      }
      Dfs(u, v);
      lowest |= (mask[u] & mask[v]);
      mask[v] |= mask[u];
    }
    int bit = 1 << (lowest == 0 ? 0 : 32 - __builtin_clz(lowest));
    while (mask[v] & bit) {
      bit <<= 1;
    }
    mask[v] &= ~(bit - 1);
    mask[v] |= bit;
    ans = max(ans, 31 - __builtin_clz(mask[v]));
  };
  Dfs(0, -1);
  cout << ans << '\n';
  return 0;
}
