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 int N = 1e5 + 10;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, a[N];
long double ans;
struct BIT {
  long double c[N];
  void modify(int p, long double v) {
    while (p <= n) c[p] += v, p += p & -p;
  }
  long double query(int p) {
    long double ret = 0;
    while (p) ret += c[p], p -= p & -p;
    return ret;
  }
} T1, T2;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) a[i] = gi();
  long double sum = 0;
  for (int i = 1; i <= n; i++) ans += sum * (n - i + 1) / n / (n + 1), sum += i;
  for (int i = n; i; i--) {
    ans += T1.query(a[i]);
    ans -= T1.query(a[i]) * 2 * i / n;
    ans += T2.query(a[i]) * 2 * i / n / (n + 1);
    T1.modify(a[i], 1);
    T2.modify(a[i], i);
  }
  cout << setprecision(10) << fixed << ans;
  return 0;
}