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 = 101000;
long long bit[N], sm[N];
int n, lin[N], a[N];
double ans = 0, rev;
inline bool cmp(int x, int y) { return a[x] > a[y]; }
inline void modify(long long *a, int x, int v) {
  for (; x <= n; x += x & -x) a[x] += v;
}
inline long long query(long long *a, int x) {
  long long tmp = 0;
  for (; x; x -= x & -x) tmp += a[x];
  return tmp;
}
int main() {
  ios ::sync_with_stdio(false);
  cin >> n;
  int x;
  long long tmp, rec;
  for (int i = 1, nn = int(n); i <= nn; ++i) cin >> a[i], lin[i] = i;
  sort(lin + 1, lin + 1 + n, cmp);
  for (int i = 2, nn = int(n); i <= nn; ++i)
    ans += (double)(n - i + 1) * (double)i * (double)(i - 1) / (double)4;
  for (int i = 1, nn = int(n); i <= nn; ++i) {
    x = lin[i];
    tmp = query(bit, x), rec = query(sm, x);
    rev += rec, ans -= (double)(n - x + 1) * (double)tmp;
    modify(bit, x, x), modify(sm, x, 1);
  }
  ans += (double)rev * (double)n * (double)(n + 1) / (double)2;
  ans /= (double)n * (double)(n + 1) / (double)2;
  cout.setf(ios ::fixed);
  cout << setprecision(12) << ans << endl;
  return 0;
}