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 maxn = 100000 + 5;
int a[maxn];
long long tree[maxn];
int n;
void add(int x, int y) {
  for (int i = x; i <= n; i += (i & (-i))) tree[i] += y;
}
long long get(int x) {
  long long ret = 0;
  for (int i = x; i; i -= i & (-i)) ret += tree[i];
  return ret;
}
int main() {
  scanf("%d", &n);
  double ans = 0;
  double tmp = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    long long cnt = (i - get(a[i]));
    ans += cnt;
    add(a[i], 1);
    long long l = i + 1;
    tmp += (l * (l - 1) * (n - l + 1) / 4.0);
  }
  for (int i = 1; i <= n; i++) tree[i] = 0;
  for (int i = 0; i < n; i++) {
    long long cnt = get(n) - get(a[i]);
    tmp -= cnt * (n - i);
    add(a[i], i + 1);
  }
  tmp = tmp / (1.0 * n * (n + 1) / 2);
  printf("%.11lf\n", ans + tmp);
}