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 MAXN = 501 * 1001;
long long fen[2][MAXN];
void add(long long x, long long v, long long id) {
  for (; x < MAXN; x += x & -x) fen[id][x] += v;
}
long long get(long long x, long long id) {
  long long ans = 0;
  for (; x; x -= x & -x) ans += fen[id][x];
  return ans;
}
long long a[MAXN];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  double ans = 0;
  cin >> n;
  double N = n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    long long x = get(a[i], 0);
    ans += (double)(n - i) * x / n / (n + 1);
    long long t = get(n + 1, 1) - get(a[i], 1);
    long long y = get(n + 1, 0) - get(a[i], 0);
    ans += t - (double)(n - i) * y / n / (n + 1);
    add(a[i], i + 1, 0);
    add(a[i], 1, 1);
  }
  cout << fixed << setprecision(9) << ans;
}