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 = 100005;
int n, a[N], b[N];
void upd(int x) {
  for (; x; x -= x & -x) b[x]++;
}
int qry(int x) {
  int y = 0;
  for (; x <= n; x += x & -x) y += b[x];
  return y;
}
long long t[N];
double ans, f[N];
void Upd(int x, long long y) {
  for (; x; x -= x & -x) t[x] += y;
}
long long Qry(int x) {
  long long y = 0;
  for (; x <= n; x += x & -x) y += t[x];
  return y;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) ans += qry(a[i]), upd(a[i]);
  long long S = 1ll * (n + 1) * n / 2;
  for (int i = 1; i <= n; ++i) {
    ans -= (double)(n - i + 1) * Qry(a[i]) / S;
    Upd(a[i], i);
  }
  for (int i = 1; i <= n; ++i) f[i] = f[i - 1] + (double)(i - 1) / 2;
  for (int i = 1; i <= n; ++i) ans += f[i] * (n - i + 1) / S;
  printf("%.9f\n", ans);
  return 0;
}