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;
template <typename T>
inline void read(T &x) {
  x = 0;
  T fl = 1;
  char ch = 0;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') fl = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + (ch ^ 48);
  x *= fl;
}
double ans = 0;
long long n, a[100005], d[100005], c[100005];
int main() {
  read(n);
  if (n == 1) return printf("0\n"), 0;
  for (int i = (1); i <= (n); i++) read(a[i]);
  for (int i = n; i; i--) {
    for (int j = a[i]; j; j -= (j & -j))
      ans += c[j], ans -= 2.0 * i * d[j] / n / (n + 1);
    for (int j = a[i]; j <= n; j += (j & -j)) c[j]++, d[j] += n - i + 1;
    ans += (n - i + 1.0) * i * (i - 1) / n / (n + 1) / 2;
  }
  printf("%.12f\n", ans);
  return 0;
}