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;

int n, answer;
vector < int > G[100020];
int A[100020][30];

void go(int v, int p = 0)
{
  for(int i = 0; i < G[v].size(); ++ i)
  {
    int to = G[v][i];
    if(to == p)
      continue;
    go(to, v);
    for(int k = 0; k < 30; ++ k)
      A[v][k] += A[to][k];
  }
  int x = 0;
  for(int i = 29; i >= 0; -- i)
    if(A[v][i] > 1)
    {
      x = i; break;
    }
  while(A[v][x])
    x ++;
  answer = max(x, answer);
  for(int i = 0; i < x; ++ i)
    A[v][i] = 0;
  A[v][x] = 1;
  for(int i = x; i < 30; ++ i)
    if(A[v][i])
      A[v][i] = 1;
}

int main()
{
  scanf("%d", &n);
  for(int i = 1; i < n; ++ i)
  {
    int x, y;
    scanf("%d %d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }

  go(1);

  printf("%d\n", answer);
}
