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 = 100010;
double ans;
int n, a[N];
pair<int, long long> operator+(pair<int, long long> a, pair<int, long long> b) {
  return pair<int, long long>(a.first + b.first, a.second + b.second);
}
void operator+=(pair<int, long long> &a, pair<int, long long> b) {
  a.first += b.first;
  a.second += b.second;
}
pair<int, long long> t[N];
void upd(int x, pair<int, long long> v) {
  for (; x > 0; x -= x & -x) t[x] += v;
}
pair<int, long long> get(int x) {
  pair<int, long long> res = pair<int, long long>(0, 0);
  for (; x <= n; x += x & -x) res += t[x];
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    pair<int, long long> res = get(a[i]);
    ans += res.first;
    ans -= 2.0 * ((double)(n - i + 1) / n) * ((double)res.second / (n + 1));
    upd(a[i], pair<int, long long>(1, i));
  }
  for (int i = 1; i <= n; ++i)
    ans += 0.5 * ((double)(1ll * i * (i - 1)) / (n + 1)) *
           ((double)(n - i + 1) / n);
  cout << setprecision(9) << fixed << ans << endl;
}