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;
double n;
int l, d, j;
double suma = 0.0;
typedef struct niz {
  int vrednost;
  double pozicija;
} niz;
niz a[100001], c[100001];
void merge_count1(int x, int y) {
  int m;
  m = ((x + y) / 2);
  if (x < y) {
    merge_count1(x, m);
    merge_count1(m + 1, y);
    l = x;
    d = m + 1;
    j = x;
    double suma_indeksa = 0.0;
    double broj_indeksa = 0.0;
    while (l <= m && d <= y) {
      if (a[l].vrednost < a[d].vrednost) {
        c[j].vrednost = a[l].vrednost;
        c[j].pozicija = a[l].pozicija;
        suma_indeksa += a[l].pozicija;
        broj_indeksa = broj_indeksa + 1.0;
        j++;
        l++;
      } else {
        c[j].vrednost = a[d].vrednost;
        c[j].pozicija = a[d].pozicija;
        suma += (broj_indeksa * n * (n + 1.0) / 2.0);
        suma -= (suma_indeksa * (n - a[d].pozicija + 1.0));
        j++;
        d++;
      }
    }
    while (d <= y) {
      c[j].vrednost = a[d].vrednost;
      c[j].pozicija = a[d].pozicija;
      suma += (broj_indeksa * n * (n + 1.0) / 2.0);
      suma -= (suma_indeksa * (n - a[d].pozicija + 1.0));
      j++;
      d++;
    }
    while (l <= m) {
      c[j].vrednost = a[l].vrednost;
      c[j].pozicija = a[l].pozicija;
      j++;
      l++;
    }
    for (int i = x; i <= y; i++) {
      a[i].vrednost = c[i].vrednost;
      a[i].pozicija = c[i].pozicija;
    }
  }
}
int main() {
  scanf("%lf", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].vrednost);
    a[i].vrednost = ((int)n) - a[i].vrednost + 1;
    a[i].pozicija = (double)i;
  }
  for (double i = 1.0; i <= n; i++) suma += (i * (i - 1.0) * (n - i + 1.0) / 4);
  merge_count1(1, (int)n);
  suma /= (n * (n + 1.0) / 2.0);
  printf("%.12lf", suma);
}