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 = 100005;
long long hjj[4 * MAXN];
long long hj[4 * MAXN];
long long sum(long long *t, long long v, long long tl, long long tr,
              long long l, long long r) {
  if (l > r) return 0LL;
  if (l == tl && r == tr) return (long long)t[v];
  long long tm = (tl + tr) / 2;
  return sum(t, v * 2, tl, tm, l, min(r, tm)) +
         sum(t, v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
void upd(long long *t, long long v, long long tl, long long tr, long long pos,
         long long x) {
  if (tl == tr) {
    t[v] = x;
    return;
  }
  long long tm = (tl + tr) / 2;
  if (pos <= tm)
    upd(t, v * 2, tl, tm, pos, x);
  else
    upd(t, v * 2 + 1, tm + 1, tr, pos, x);
  t[v] = t[v * 2] + t[v * 2 + 1];
}
int main() {
  long long n;
  long long inv_total = 0;
  long long inv = 0;
  long long d = 0;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long p;
    cin >> p;
    p--;
    d += sum(hj, 1, 0, n - 1, p + 1, n - 1);
    inv_total += d;
    inv += sum(hjj, 1, 0, n - 1, p + 1, n - 1);
    upd(hj, 1, 0, n - 1, p, i);
    upd(hjj, 1, 0, n - 1, p, 1);
  }
  double randi = 0;
  double rand_total = 0;
  for (long long i = 2; i <= n; i++) {
    randi += (double)(i - 1) / 2;
    rand_total += randi * (double)(n - i + 1);
  }
  double ans = inv + (rand_total - (double)inv_total) /
                         (double)((long long)n * (long long)(n + 1) / 2);
  cout << setprecision(16) << fixed << ans;
  return 0;
}