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>
int inp() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum;
}
struct BIT {
  long long f[100010];
  void modify(int x, int c) {
    for (int i = x; i <= 100000; i += ((i) & (-i))) f[i] += c;
  }
  long long query(int x) {
    long long sum = 0;
    for (int i = x; i; i -= ((i) & (-i))) sum += f[i];
    return sum;
  }
  void clear() { memset(f, 0, sizeof(f)); }
} t;
int a[100010];
int main() {
  int n = inp();
  double ans = 0;
  double tot = (double)(n) * (double)(n + 1) / 2.0;
  t.clear();
  for (int i = 1; i <= n; i++) {
    a[i] = inp();
    t.modify(a[i], 1);
    ans += (t.query(n) - t.query(a[i]));
  }
  t.clear();
  for (int i = n; i >= 1; i--) {
    ans -= (double)(t.query(a[i] - 1) * i) / tot;
    t.modify(a[i], n - i + 1);
  }
  for (int i = 2; i <= n; i++)
    ans +=
        (double)((long long)(n - i + 1) * (long long)(i) * (long long)(i - 1)) /
        (tot * 4.0);
  printf("%.20lf\n", ans);
}