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>

enum {
  MAXN = 100000 + 5
};

typedef std::vector<int> vi;

int n;
vi e[MAXN];
int msk[MAXN];

void dfs(int u, int pa) {
  int d, c;
  for (int v: e[u]) if (v != pa) dfs(v, u);
  for (d=20; ~d; --d) {
    c = 0;
    for (int v: e[u]) if (v != pa && msk[v]>>d&1) ++c;
    if (c >= 2) {
      msk[u] += (1<<(d+1));
      break;
    } else if (c == 1)
      msk[u] |= 1<<d;
  }
  if (d == -1) ++msk[u];
}

void clear() {
  memset(msk, 0, sizeof msk);
}

void exec() {
  int i, u, v, ans;
  clear();
  scanf("%d", &n);
  for (i=0; i<n-1; ++i) {
    scanf("%d%d", &u, &v), --u, --v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(0, -1);
  ans = 0;
  while ((1<<(ans+1)) <= msk[0]) ++ans;
  printf("%d\n", ans);
}

int main() {
  if (fopen("D.in", "r") != NULL) {
    freopen("D.in", "r", stdin);
    freopen("D.out", "w", stdout);
  }
  exec();
  return 0;
}
