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 INF = 1000 * 1000 * 1000;
const long long LINF = INF * (long long)INF;
const int mod = 1000 * 1000 * 1000 + 7;
const int MAX = 100005;
const int LG = 20;
long long ft[2][MAX];
void inc(int idx, int pos, int delta) {
  for (; pos < MAX; pos |= pos + 1) ft[idx][pos] += delta;
}
long long sum(int idx, int pos) {
  long long res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) res += ft[idx][pos];
  return res;
}
int pos[MAX];
int main() {
  double ans = 0;
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) ans += i * (i - 1) * (n - i + 1) / 4.;
  long long div = n * (n + 1) / 2;
  for (int i = 0; i < n; i++) {
    int first;
    cin >> first;
    first--;
    pos[first] = i;
  }
  for (int i = n - 1; i >= 0; i--) {
    ans += sum(0, pos[i]) * n * (n + 1) / 2;
    ans -= (n - pos[i]) * sum(1, pos[i]);
    inc(0, pos[i], 1);
    inc(1, pos[i], pos[i] + 1);
  }
  cout << fixed << setprecision(10) << ans / div << endl;
}