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;

const int N = 123456;

int n, foo[N], key[N], answer;
vector<int> adj[N];

void dfs(int x, int f) {
  int bar = 0;
  for (auto y : adj[x]) {
    if (y != f) {
      dfs(y, x);
      bar |= foo[x] & foo[y];
      foo[x] |= foo[y];
    }
  }
  if (!foo[x]) {
    key[x] = 0;
    foo[x] = 1;
  } else {
    while ((1 << key[x]) <= bar || (foo[x] >> key[x] & 1)) {
      ++key[x];
    }
    for (int i = 0; i < key[x]; ++i) {
      foo[x] -= (foo[x] >> i & 1) << i;
    }
    foo[x] |= 1 << key[x];
    answer = max(answer, key[x]);
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1, x, y; i < n; ++i) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  cout << answer << '\n';
  return 0;
}