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 = 100004;
long long bit[N];
int a[N];
void update(int x, int val) {
  while (x < N) {
    bit[x] += val;
    x += x & -x;
  }
  return;
}
long long query(int x) {
  long long ans = 0;
  while (x > 0) {
    ans = ans + bit[x];
    x -= x & -x;
  }
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  int i;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  double totl = (double)(1LL * n * (n + 1)) / 2.0;
  double ans = 0.0;
  double lensum = 0.0;
  double ssum = 0.0;
  for (i = n; i >= 1; i--) {
    ssum = ssum + (double)query(a[i]);
    lensum = lensum + (1LL * (n - i) * (n - i + 1)) / 4.0;
    ans = ans + (lensum - ssum) / totl;
    update(a[i], n - i + 1);
  }
  memset(bit, 0, sizeof(bit));
  long long cnt = 0;
  for (i = n; i > 0; i--) {
    cnt = cnt + query(a[i]);
    update(a[i], 1);
  }
  ans = ans + cnt;
  printf("%0.20lf\n", ans);
  return 0;
}