Problem: You are given a permutation of integers from 1 to n. Exactly once you apply the following operation to this permutation: pick a random segment and shuffle its elements. Formally:

  1. Pick a random segment (continuous subsequence) from l to r. All <image> segments are equiprobable. 
  2. Let k = r - l + 1, i.e. the length of the chosen segment. Pick a random permutation of integers from 1 to k, p1, p2, ..., pk. All k! permutation are equiprobable. 
  3. This permutation is applied to elements of the chosen segment, i.e. permutation a1, a2, ..., al - 1, al, al + 1, ..., ar - 1, ar, ar + 1, ..., an is transformed to a1, a2, ..., al - 1, al - 1 + p1, al - 1 + p2, ..., al - 1 + pk - 1, al - 1 + pk, ar + 1, ..., an. 



Inversion if a pair of elements (not necessary neighbouring) with the wrong relative order. In other words, the number of inversion is equal to the number of pairs (i, j) such that i < j and ai > aj. Find the expected number of inversions after we apply exactly one operation mentioned above.

Input

The first line contains a single integer n (1 ≤ n ≤ 100 000) — the length of the permutation.

The second line contains n distinct integers from 1 to n — elements of the permutation.

Output

Print one real value — the expected number of inversions. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 9. 

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.

Example

Input

3
2 3 1


Output

1.916666666666666666666666666667

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n + 1);
  vector<long long> ft(n + 10, 0);
  auto update = [&](int pos, int val) {
    while (pos < int(ft.size())) {
      ft[pos] += val;
      pos += pos & -pos;
    }
  };
  auto get = [&](int pos) {
    long long ret = 0;
    while (pos) {
      ret += ft[pos];
      pos -= pos & -pos;
    }
    return ret;
  };
  long long inv = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    inv += i - 1 - get(a[i]);
    update(a[i], 1);
  }
  fill(begin(ft), end(ft), 0);
  double iden = 1.0 / (1LL * n * (n + 1));
  double diff = 0;
  for (int i = n; i >= 1; i--) {
    diff += iden * (get(n) - get(a[i])) * i;
    diff -= iden * get(a[i]) * i;
    update(a[i], n - i + 1);
  }
  cout << fixed << setprecision(15) << inv + diff << '\n';
  return (0);
}