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 + 5;
int a[N], n;
long long Bit[N];
long double F[N], ans;
void update(int u, int p) {
  for (; u < N; u += u & -u) Bit[u] += 1LL * p;
}
long long get(int u) {
  long long res = 0;
  for (; u > 0; u -= u & -u) res += Bit[u];
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) F[i] = F[i - 1] + (i - 1.) / 2.;
  for (int i = n; i >= 1; --i) {
    ans += 1. * get(a[i]);
    update(a[i], 1);
  }
  memset(Bit, 0, sizeof Bit);
  for (int i = n; i >= 1; --i) {
    long double res = get(a[i]);
    res = 2. * res * i / n / (n + 1);
    ans -= res;
    update(a[i], n - i + 1);
  }
  for (int i = 1; i <= n; ++i)
    ans = ans + 2. * (F[i] * (n - i + 1)) / n / (n + 1);
  cout << fixed << setprecision(9) << ans;
}