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 long long MXN = 3e5 + 10;
long long n, inv;
long double ans, t;
long long A[MXN], Fen[MXN];
void Upd(long long p, long long x) {
  for (; p < MXN; p += p & -p) Fen[p] += x;
}
long long Get(long long p) {
  long long s = 0;
  for (; p; p -= p & -p) s += Fen[p];
  return s;
}
long long Get(long long l, long long r) { return Get(r) - Get(l - 1); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n, t = 1.0 * n * (n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    inv += Get(A[i], n);
    Upd(A[i], +1);
  }
  memset(Fen, 0, sizeof Fen);
  long double ea = 0.0, er = 0.0;
  for (int i = 1; i <= n; i++) {
    ea += 1.0 * Get(1, A[i]) * (n - i + 1.0) / t;
    er += 1.0 * Get(A[i], n) * (n - i + 1.0) / t;
    Upd(A[i], i);
  }
  cout << fixed << setprecision(13);
  ans = inv + ea - er;
  cout << ans << '\n';
  return 0;
}