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>
const int N = 100010;
int n, i, x, p[N];
double ans, bit[N];
inline void add(int x, double p) {
  for (; x <= n; x += x & -x) bit[x] += p;
}
inline double ask(int x) {
  double t = 0;
  for (; x; x -= x & -x) t += bit[x];
  return t;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &x), p[x] = i;
  for (i = 1; i <= n; i++) {
    ans += 1.0 * (n - p[i] + 1) / n / (n + 1) * ask(p[i]);
    add(p[i], p[i]);
  }
  for (i = 1; i <= n; i++) bit[i] = 0;
  for (i = n; i; i--) {
    ans -= 1.0 * (n - p[i] + 1) / n / (n + 1) * ask(p[i]);
    add(p[i], p[i]);
  }
  for (i = 1; i <= n; i++) bit[i] = 0;
  for (i = n; i; i--) {
    ans += ask(p[i]);
    add(p[i], 1);
  }
  printf("%.10f", ans);
}