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;
struct Bit {
  double a[100002];
  int n;
  void add(int x, double val) {
    for (; x <= n; x += x & -x) a[x] += val;
  }
  double que(int x) {
    double res = 0;
    for (; x > 0; x -= x & -x) res += a[x];
    return res;
  }
} bit, bit2;
double a[100002], ans;
int main() {
  int n;
  scanf("%d", &n);
  bit.n = bit2.n = n;
  for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
  for (int i = 1; i <= n; i++) {
    ans += bit2.que(n - a[i] + 1) * n * (n + 1) / 2;
    bit2.add(n - a[i] + 1, 1);
    ans += (double)i * (i - 1) / 4 * (n - i + 1);
  }
  for (int i = 1; i <= n; i++) {
    ans -= (n - i + 1) * bit.que(n - a[i] + 1);
    bit.add(n - a[i] + 1, i);
  }
  ans /= (double)n * (n + 1) / 2;
  printf("%.12f\n", ans);
}