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;
long long int powmod(long long int a, long long int e) {
  if (e == 0) return 1;
  long long int x = powmod(a * a % 1000000007, e >> 1);
  return e & 1 ? x * a % 1000000007 : x;
}
long long int a[(long long int)100001];
double pref[(long long int)100001];
long long int n;
long long int fen[(long long int)100001];
long long int fenw[(long long int)100001];
void update(long long int index, long long int up, long long int fen[]) {
  while (index <= n) {
    fen[index] += up;
    index += (index & -index);
  }
}
long long int sum(long long int index, long long int fen[]) {
  long long int z = 0;
  while (index) {
    z += (fen[index]);
    index -= (index & -index);
  }
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (long long int i = 0; i < n; i++) cin >> a[i + 1];
  cout.precision(11);
  for (long long int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1] + 1.0 * (i * (i - 1)) / 4.0;
  }
  long long int inv = 0;
  double ans = 0;
  long long int total = (n * (n + 1)) / 2;
  long long int tinv = 0;
  for (long long int i = n; i; i--) {
    tinv += sum(a[i], fenw);
    inv += sum(a[i], fen);
    ans += 1.0 * (pref[n + 1 - i] - inv) / total;
    update(a[i], n - i + 1, fen);
    update(a[i], 1, fenw);
  }
  cout << tinv + ans;
}