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;
typedef long long ll;

const int MaxN = 100010;
const int MaxLabel = 20;

int n, ans, f[MaxN]; vector<int> edge[MaxN];

void dfs(int x, int fa) {
  int all = 0;
  for (int y : edge[x]) {
    if (y == fa)
      continue;
    dfs(y, x);
    all |= (f[x] & f[y]);
    f[x] |= f[y];
  }
  
  int t = 0;
  while ((1 << t) <= all || f[x] >> t & 1)
    ++t;
  ans = max(ans, t);
  f[x] >>= t;
  f[x] <<= t;
  f[x] |= (1 << t);
}

int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  
  dfs(1, 0);
  cout << ans << "\n";
  return 0;
}

