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;
using ll = long long;
int N, a[100100];
ll totgood, good;
struct tree {
  ll fen[100100];
  void u(int i, ll v) {
    while (i <= N) fen[i] += v, i += i & (-i);
  }
  ll q(int i) {
    ll r = 0;
    while (i >= 1) r += fen[i], i -= i & (-i);
    return r;
  }
} g, k;
int main() {
  scanf("%d", &N);
  for (int i = (0); i < (N); i++) {
    scanf("%d", &a[i]);
    good += g.q(a[i] - 1), g.u(a[i], 1);
    totgood += k.q(a[i] - 1) * (N - i), k.u(a[i], i + 1);
  }
  long double n = N;
  long double dv = n * (n + 1.);
  long double v = (n * (n - 1) / 2. - ((long double)good)) * dv;
  long double f = (n - 1) * n * (n + 1) * (n + 2) / 24.;
  long double t = ((long double)totgood * 2) - f;
  printf("%.16lf\n", double((v + t) / dv));
  return 0;
}