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 MAXN = 100 * 1000 + 17;
long long fen[MAXN][2], a[MAXN];
long double ans, res, inv, n;
long long get(bool flag, int x) {
  long long res = 0;
  for (; x; x -= x & (-x)) res += fen[x][flag];
  return res;
}
void add(int x, long long y) {
  for (; x <= n; x += x & (-x)) fen[x][0]++, fen[x][1] += y;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (long long i = n; i > 0; i--) {
    inv += get(0, a[i]);
    res += i * get(1, a[i]);
    add(a[i], n - i + 1);
  }
  res = -res;
  for (long long i = 2; i <= n; i++) res += (n - i + 1) * i * (i - 1) / 4;
  ans = inv;
  ans += res / (n * (n + 1) / 2);
  cout << fixed << setprecision(11) << ans << endl;
  return 0;
}