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 mod = (int)1e9 + 7;
const int inf = (1 << 30) - 1;
const int N = (int)2e5 + 5;
int n, a[N];
long long f[N], invs;
void init() {
  for (int i = 1; i <= n; i++) f[i] = 0;
}
void upd(int R, int x) {
  while (R <= n) {
    f[R] += x;
    R |= R + 1;
  }
}
long long get(int R) {
  long long res = 0;
  while (R > 0) {
    res += f[R];
    R = (R & (R + 1)) - 1;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    invs += i - 1 - get(a[i]);
    upd(a[i], 1);
  }
  init();
  double ans = 0;
  for (int i = 1; i <= n; i++) {
    long long sumLess = get(a[i]);
    long long sumMore = get(n) - get(a[i]);
    ans += (n + 1 - i) * 1.0 / n * (sumLess / (n + 1.0));
    ans -= (n + 1 - i) * 1.0 / n * (sumMore / (n + 1.0));
    upd(a[i], i);
  }
  printf("%.12f\n", invs + ans);
  return 0;
}