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;
int n, a[1000010];
long long c[1000010], s[1000010];
inline void add(long long *c, int x, int v) {
  for (; x <= n; x += ((x) & -(x))) c[x] += v;
}
inline long long ask(long long *c, int x) {
  long long res = 0;
  for (; x; x -= ((x) & -(x))) res += c[x];
  return res;
}
int main() {
  scanf("%d", &n);
  int i;
  long double ans = 0, res = 0;
  for (i = 1; i <= n; i++) scanf("%d", a + i);
  memset(c, 0, sizeof(c)), memset(s, 0, sizeof(s));
  for (i = 1; i <= n; i++) res += (long double)i * (i - 1) * (n - i + 1);
  ans += res / 2.0 / n / (n + 1);
  res = 0;
  for (i = n; i; i--) {
    ans += (long double)ask(c, a[i] - 1), add(c, a[i], 1);
    res -= (long double)ask(s, a[i] - 1) * i, add(s, a[i], n - i + 1);
  }
  ans += res * 2.0 / n / (n + 1);
  printf("%.20f", (double)ans);
  return 0;
}