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, a[100010];
double dp[100010], tot = 0.0, sum = 0.0, C[100010], Cn[100010];
void update(int x, double v) {
  for (register int i = x; i <= n; i += (i & (-i))) C[i] += v;
}
double query(int x) {
  double res = 0.0;
  for (register int i = x; i; i -= (i & (-i))) res += C[i];
  return res;
}
void UPDATE(int x, double v) {
  for (register int i = x; i <= n; i += (i & (-i))) Cn[i] += v;
}
double QUERY(int x) {
  double res = 0.0;
  for (register int i = x; i; i -= (i & (-i))) res += Cn[i];
  return res;
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  dp[1] = 0.0;
  for (register int i = 2; i <= n; ++i)
    dp[i] = dp[i - 1] + (1.0 * i - 1.0) / 2.0;
  for (register int i = n; i >= 1; --i) {
    tot += QUERY(a[i] - 1);
    UPDATE(a[i], 1.0);
    sum += i * query(a[i] - 1);
    update(a[i], n * 1.0 - i * 1.0 + 1.0);
  }
  sum /= (n * 1.0 * (n * 1.0 + 1.0) / 2.0);
  tot -= sum;
  for (register int i = 2; i <= n; ++i)
    tot +=
        dp[i] * (n * 1.0 - i * 1.0 + 1.0) / (n * 1.0 * (n * 1.0 + 1.0) / 2.0);
  printf("%.15f\n", tot + (1e-15));
  return 0;
}