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;
using pii = pair<int, int>;
using ppi = pair<pii, int>;
using vi = vector<int>;
int N = 5e5 + 10;
vector<ld> B1, B2;
void add(vector<ld>& B, int i, int x) {
  int n = B.size();
  for (; i < n; i += i & (-i)) B[i] = B[i] + x;
}
ld sum(vector<ld>& B, int i) {
  ld sum = 0;
  for (; i > 0; i -= i & (-i)) sum = sum + B[i];
  return sum;
}
void solve() {
  int n;
  cin >> n;
  B1.assign(n + 5, 0);
  B2.assign(n + 5, 0);
  vector<int> arr(n);
  for (int& e : arr) cin >> e;
  ll sum1 = 0, sum2 = 0;
  ld fin = 0;
  ld tot = (1LL * n * (n + 1)) / 2;
  for (int i = n - 1; i >= 0; i--) {
    int c = arr[i];
    sum1 += sum(B1, c);
    sum2 += sum(B2, c);
    fin += sum2;
    add(B1, c, 1);
    add(B2, c, n - i);
  }
  ld ans = sum1 - fin / tot + (tot - 1) / 12;
  cout << fixed << setprecision(30) << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) solve();
}