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;
using ll = long long;
using ld = long double;
int const nmax = 100000;
class FenwickTree {
 private:
  int n;
  vector<ld> aib;

 public:
  FenwickTree(int n) {
    this->n = n;
    aib.resize(1 + n);
  }
  void update(int x, int val) {
    while (x <= n) {
      aib[x] += val;
      x += (x ^ (x & (x - 1)));
    }
  }
  ld query(int x) {
    ll result = 0;
    while (0 < x) {
      result += aib[x];
      x = (x & (x - 1));
    }
    return result;
  }
};
int v[1 + nmax];
int n;
int main() {
  cin >> n;
  ld total = 1LL * n * (n + 1) / 2;
  for (int i = 1; i <= n; i++) cin >> v[i];
  FenwickTree aib(n), aib2(n);
  ld result = 0, sum = 0;
  for (int i = 1; i <= n; i++) {
    ld cand = aib2.query(n) - aib2.query(v[i]);
    result += (total * cand - (aib.query(n) - aib.query(v[i])) * (n - i + 1.0));
    result += aib.query(n) * (n - i + 1.0) / 2.0;
    aib.update(v[i], i);
    aib2.update(v[i], 1);
  }
  cout << setprecision(13) << fixed << result / total;
  return 0;
}