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;
vector<long long> f1, f2, f3;
long long query(long long i, vector<long long> &f) {
  double sum = 0;
  while (i > 0) {
    sum += f[i];
    i -= (i & (-i));
  }
  return sum;
}
void upd(long long i, long long k, vector<long long> &f) {
  while (i < f.size()) {
    f[i] += k;
    i += (i & (-i));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  long long n;
  cin >> n;
  long double sum = 0.;
  vector<long long> v(n);
  vector<long long> counter(n, 0);
  f1 = f2 = f3 = vector<long long>(n + 1, 0);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    upd(v[i], 1, f1);
    upd(v[i], i, f2);
    upd(v[i], i * i, f3);
    long double v1 = query(n, f1) - query(v[i], f1),
                v2 = query(n, f2) - query(v[i], f2),
                v3 = query(n, f3) - query(v[i], f3);
    long double j = i;
    sum +=
        v1 * j * (j + 1) / (n * (n + 1)) +
        (v1 * n * n - 2 * n * v2 + v3 - v1 * n + v2) / (n * (n + 1)) -
        (1.L / (n * (n + 1)) * (v1 * j * j - 2.L * j * v2 + v3 - v1 * j + v2));
  }
  for (int i = 1; i <= n; ++i) {
    sum += ((long double)(n + 1 - i)) * i * (i - 1) / 4. / (n * (n + 1) / 2);
  }
  cout.setf(ios::fixed);
  cout.precision(15);
  cout << sum << endl;
}