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 = 1e5 + 10;
int a[N];
long long s[N], t[N];
void add(int x, int ind, long long *arr) {
  for (int i = ind; i < N; i += i & -i) arr[i] += x;
}
long long get(int ind, long long *arr) {
  long long ans = 0;
  for (int i = ind; i > 0; i -= i & -i) ans += arr[i];
  return ans;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long double ans = 0;
  for (int i = 1; i <= n; i++)
    ans += 0.5 * (n + 1 - i) * i * (i - 1) / (n + 1) / n;
  for (int i = 0; i < n; i++) {
    long long cnt = get(n, t) - get(a[i], t);
    long long sum = get(n, s) - get(a[i], s);
    ans += 1.0 * cnt;
    ans -= 2.0 * (n - i) * sum / (n + 1) / n;
    add(1, a[i], t);
    add(i + 1, a[i], s);
  }
  cout << fixed << setprecision(16) << ans << endl;
}