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 + 7;
int a[N];
inline int lowbit(int x) { return x & -x; }
struct BIT {
  long long c[N];
  void modify(int x, int v) {
    for (int i = x; i < N; i += lowbit(i)) c[i] += v;
  }
  long long query(int x) {
    long long res = 0;
    for (int i = x; i; i -= lowbit(i)) res += c[i];
    return res;
  }
} c, s;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  double res = 0;
  for (int i = 1; i <= n; ++i) {
    double cur = i * 0.5 * (n - i) * (n - i + 1);
    res += cur;
  }
  for (int i = n; i >= 1; --i) {
    double pres =
        c.query(a[i] - 1) * 1.0 * (n * 1.0 * n + n - 2 * i - 2.0 * n * i);
    pres += 2.0 * i * (s.query(a[i] - 1));
    c.modify(a[i], 1);
    s.modify(a[i], i);
    res += pres;
  }
  printf("%.18lf\n", res / n / (n + 1));
  return 0;
}