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;
int n, i, j, x, v[100005];
double aib[3][100005];
inline int lsb(int x) { return x & (-x); }
void update(int idx, int position, double val) {
  for (i = position; i < 100005; i += lsb(i)) aib[idx][i] += val;
}
double query(int idx, int pos) {
  double res = 0;
  for (int i = pos; i > 0; i -= lsb(i)) res += aib[idx][i];
  return res;
}
int main() {
  scanf("%d", &n);
  double sol = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
  }
  for (i = 1; i <= n; i++) sol += 1LL * i * (i - 1) * (n + 1 - i) / 4.0;
  for (int i = n; i >= 1; i--) {
    sol += (double)query(1, v[i]) *
           (1LL * (n - i) * (n - i + 1) / 2 + 1LL * (i - 1) * i / 2);
    sol += (double)(query(2, v[i]) - query(1, v[i]) * i) * i;
    update(1, v[i], 1);
    update(2, v[i], i);
  }
  printf("%.20f", (double)sol / n / (n + 1) * 2);
  return 0;
}