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;
long long N, a[100010], s[100100], cnt[100010];
void update(long long *t, long long pos, long long val) {
  for (; pos <= N; pos += pos & (-pos)) t[pos] += val;
}
long long query(long long *t, long long r) {
  long long res = 0;
  for (; r > 0; r -= r & (-r)) res += t[r];
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  int i;
  long long nr, sum;
  long double res = 0;
  for (i = 1; i <= N; i++) {
    cin >> a[i];
    update(cnt, a[i], 1);
    update(s, a[i], i);
  }
  for (i = 1; i <= N; i++) {
    nr = query(cnt, a[i] - 1);
    sum = query(s, a[i] - 1);
    res += nr - (1.0 * i * (nr * (N + 1) - sum)) / (N * (N + 1));
    nr = N - i - nr;
    sum = query(s, N) - sum - i;
    res += (1.0 * i * (nr * (N + 1) - sum)) / (N * (N + 1));
    update(cnt, a[i], -1);
    update(s, a[i], -i);
  }
  cout << fixed << setprecision(15) << res << "\n";
  return 0;
}