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>
const long long mod = 1000000007ll;
using namespace std;
struct FT {
  vector<long long> s;
  FT(long long n) : s(n) {}
  void update(long long pos, long long dif) {
    for (; pos < ((long long)(s).size()); pos |= pos + 1) s[pos] += dif;
  }
  long long query(long long pos) {
    long long res = 0;
    for (; pos > 0; pos &= pos - 1) res += s[pos - 1];
    return res;
  }
  long long range(long long l, long long r) { return query(r + 1) - query(l); }
  long long lower_bound(long long sum) {
    if (sum <= 0) return -1;
    long long pos = 0;
    for (long long pw = 1 << 25; pw; pw >>= 1) {
      if (pos + pw <= ((long long)(s).size()) && s[pos + pw - 1] < sum)
        pos += pw, sum -= s[pos - 1];
    }
    return pos;
  }
};
int main() {
  cin.sync_with_stdio(0);
  long long n;
  cin >> n;
  vector<long long> perm(n);
  for (long long i = 0; i < (n); i++) {
    cin >> perm[i];
    perm[i]--;
  }
  FT ft(n);
  FT ft1s(n);
  long long invs_in_all_segs = 0;
  long long totalInvs = 0;
  for (long long i = 0; i < (n); i++) {
    invs_in_all_segs += ft.range(perm[i] + 1, n - 1) * (n - i);
    totalInvs += ft1s.range(perm[i] + 1, n - 1);
    ft.update(perm[i], i + 1);
    ft1s.update(perm[i], 1);
  }
  double totInvsAdded = 0.0;
  for (long long i = (1); i <= (n); i++) {
    totInvsAdded +=
        ((double)(i * (i - 1)) / (double)4) * ((double)(n - i + 1ll));
  }
  double avgAdded = totInvsAdded / ((double)((n * (n - 1) / 2) + n));
  double avgRem = (double)invs_in_all_segs / ((double)((n * (n - 1) / 2) + n));
  double res = (double)(totalInvs) + avgAdded - avgRem;
  cout << setprecision(15) << res << "\n";
  return 0;
}