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;
long long bit[123456] = {0}, a[123456];
long long update(long long pos, long long val) {
  while (pos < 112345) {
    bit[pos] += val;
    pos += pos & (-pos);
  }
  return 0;
}
long long query(long long pos) {
  long long sumi = 0;
  while (pos > 0) {
    sumi += bit[pos];
    pos -= pos & (-pos);
  }
  return sumi;
}
int main() {
  std::ios::sync_with_stdio(false);
  long long n, i;
  cin >> n;
  double nn = n, ans = 0, sumi = 0, sumi1 = 0, sumi2 = 0;
  nn *= 1.00;
  for (i = 1; i < n + 1; i++) {
    cin >> a[i];
  }
  double total = nn * (nn + 1);
  total /= 2;
  for (i = 1; i < n + 1; i++) {
    ans = (nn - i) * (nn - i + 1);
    ans /= 4.00;
    ans /= total;
    ans *= (i * 1.00);
    sumi += ans;
  }
  for (i = 1; i < n + 1; i++) {
    ans = query(100005) - query(a[i]);
    update(a[i], 1);
    sumi1 += ans;
  }
  sumi += (sumi1 / (nn + 1)) * (nn - 1);
  sumi += (2 * sumi1 / (nn + 1));
  for (i = 0; i < 123456; i++) {
    bit[i] = 0;
  }
  for (i = 1; i < n + 1; i++) {
    ans = query(100005) - query(a[i]);
    ans *= 1.0;
    ans /= total;
    update(a[i], i);
    sumi2 += ans * (nn - i + 1);
  }
  sumi -= sumi2;
  cout << setprecision(20) << sumi << endl;
}