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;
const int N = 100005;
long long fw[N][2];
int a[N];
long double ans = 0.0;
void add(int idx, int val, int tp) {
  for (; idx < N; idx = (idx | (idx + 1))) {
    fw[idx][tp] += val;
  }
}
long long get(int r, int tp) {
  long long res = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) {
    res += fw[r][tp];
  }
  return res;
}
int main() {
  int i, n;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int len;
  long double c0 = 0.0, c1 = 1.0, c2 = 2.0;
  for (i = 1; i <= n; i++) {
    len = i;
    ans = ans + (((c2 / ((n + c0) * (n + c1))) *
                  (((len + c0) * (len - c1) / (c2 * 2))))) *
                    (n - len + 1);
  }
  long double q = 2.0;
  for (i = 1; i <= n; i++) {
    ans += ((c2 / ((n + c0) * (n + c1))) *
            ((i - 1 - get(a[i] - 1, 0)) * ((n + c0) * (n + c1) / c2) -
             ((n - i + c1) * (((i - c1) * i / c2) - get(a[i] - 1, 1)))));
    add(a[i], i, 1);
    add(a[i], 1, 0);
  }
  cout << fixed << setprecision(20) << ans << endl;
  return 0;
}