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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAX_N = 100005;

class Edge {
public:
  int nxt, to;
} e[MAX_N << 1];

int head[MAX_N], cnt;
void addedge(int u, int v) {
  e[++cnt] = (Edge){head[u], v}, head[u] = cnt;
  e[++cnt] = (Edge){head[v], u}, head[v] = cnt;
}

int N, f[MAX_N];

void dfs(int u, int v) {
  for (int i = head[u]; i; i = e[i].nxt)
    if (e[i].to != v)
      dfs(e[i].to, u);
  int vis = 0, twice = 0;
  for (int i = head[u]; i; i = e[i].nxt) 
    if (e[i].to != v) {
      twice |= vis & f[e[i].to];
      vis |= f[e[i].to];
    }
  if (!twice) {
    f[u] = vis + 1;
  } else {
    int k = 20;
    while (!(twice >> k & 1)) k--;
    f[u] = (vis ^ (vis & ((1 << k + 1) - 1))) + (1 << k + 1);
  }
}

int main() {
  scanf("%d", &N);

  for (int i = 1, a, b; i < N; ++i) {
    scanf("%d%d", &a, &b);
    addedge(a, b);
  }

  dfs(1, 0);
  
  int k = 20;
  while (!(f[1] >> k & 1)) k--;

  printf("%d\n", k);
  return 0;
}
