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 long long SIZE = 100000000;
int n, i, x;
long long s, BIT1[100001], BIT2[100001];
double s1, s2;
void Add(long long *BIT, int x, int y) {
  for (; x <= n; x += x & -x) *(BIT + x) += (long long)y;
}
long long Sum(long long *BIT, int x) {
  long long ans = 0;
  for (; x; x -= x & -x) ans += *(BIT + x);
  return ans;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    s += Sum(BIT1, n) - Sum(BIT1, x);
    s1 += (double)(Sum(BIT2, n) - Sum(BIT2, x)) * (double)(n - i + 1) / n /
          (n + 1);
    Add(BIT1, x, 1);
    Add(BIT2, x, i);
  }
  for (i = 1; i <= n; i++)
    s2 += (s + (double)i * (double)(i - 1) / 4) * (double)(n - i + 1) / n /
          (n + 1);
  printf("%.15lf\n", (s2 - s1) * 2);
  return 0;
}