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 int bit[100002];
void update(int in, int val) {
  while (in < 100002) {
    bit[in] += val;
    in = in + (in & (-in));
  }
}
long long int get(int in) {
  long long int ans = 0;
  while (in) {
    ans += bit[in];
    in = in - (in & (-in));
  }
  return ans;
}
long long int arr[100002], tot2;
double pre[100002];
int main() {
  int n, i;
  pre[1] = 0;
  double tot = 0, rem = 0;
  for (i = 2; i < 100002; i++) {
    tot += i - 1;
    pre[i] = pre[i - 1] + tot / i;
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  for (i = n; i >= 1; i--) {
    tot2 += get(arr[i]);
    update(arr[i], 1);
  }
  for (i = 1; i <= n; i++) update(arr[i], -1);
  for (i = 1; i <= n; i++) {
    double sum = get(100002 - 1) - get(arr[i]);
    update(arr[i], i);
    sum = sum / n;
    sum = (sum * 2);
    rem = rem + sum * (n - i + 1) / (n + 1);
  }
  double ans = tot2 - rem;
  rem = 0;
  for (i = 1; i <= n; i++)
    rem = rem + (pre[i] / (n + 1)) * ((n - i + 1) * 2 / (double)n);
  ans = ans + rem;
  printf("%.10lf\n", ans);
  return 0;
}