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 <vector>

using namespace std;

const int MAX_N = 1e5 + 5;

vector<int> adj [MAX_N];
int mask [MAX_N];

void dfs (int u, int p) {
  int chb = 0;
  for (int nxt : adj[u]) {
    if (nxt != p) {
      dfs(nxt, u);
      if (mask[u] & mask[nxt]) {
        chb = max(chb, 31 - __builtin_clz(mask[u] & mask[nxt]));
      }
      mask[u] |= mask[nxt];
    }
  }

  while (mask[u] & 1 << chb) {
    chb++;
  }

  mask[u] += 1 << chb;
  mask[u] -= mask[u] % (1 << chb);
}

int main () {
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;

    adj[u].push_back(v);
    adj[v].push_back(u);
  }

  dfs(1, 0);

  cout << 31 - __builtin_clz(mask[1]) << endl;
}
