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 int N = 1e6;
double ret(double n) { return n * (n - 1) / 4.0; }
pair<long long, long long> bit[N + 2];
void upd(int id, int vl) {
  for (; id > 0; id -= id & -id) bit[id].first++, bit[id].second += vl;
}
pair<long long, long long> get(int id) {
  pair<long long, long long> ret;
  ret = {0, 0};
  for (; id <= N; id += id & -id)
    ret.first += bit[id].first, ret.second += bit[id].second;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  double ans = 0;
  for (int i = 1; i <= n; i++) ans += ret(i) * (n - i + 1);
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    pair<long long, long long> p = get(x);
    ans += p.first * ((1LL * n * (n + 1)) / 2) - (p.second * (n - i + 1));
    upd(x, i);
  }
  double d = (1LL * n * (n + 1)) / 2;
  ans /= d;
  cout << setprecision(9) << fixed << ans << endl;
  return 0;
}