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;
vector<long long> BITree;
int n;
long long getSumBIT(int index) {
  long long sum = 0;
  index = index + 1;
  while (index > 0) {
    sum += BITree[index];
    index -= index & (-index);
  }
  return sum;
}
void updateBIT(int index, int val) {
  index = index + 1;
  while (index <= n) {
    BITree[index] += val;
    index += index & (-index);
  }
}
int main() {
  scanf("%d", &n);
  vector<int> a(n);
  int i;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  double s1 = 0;
  double s2 = 0;
  double s3 = 0;
  double s4 = 0;
  double totInv = 0;
  BITree = vector<long long>(n + 1, 0);
  for (i = n - 1; i >= 0; i--) {
    if (a[i] != 0) {
      totInv += getSumBIT(a[i] - 1);
    }
    updateBIT(a[i], 1);
  }
  BITree = vector<long long>(n + 1, 0);
  double len;
  for (i = n - 1; i >= 0; i--) {
    len = n - i;
    s1 += (len * (len - 1)) / 4.0;
    s2 += s1;
    if (a[i] != 0) {
      s3 += getSumBIT(a[i] - 1);
    }
    s4 += s3;
    updateBIT(a[i], n - i);
  }
  double nums = (((double)n) * (n + 1)) / 2.0;
  double ans = (s2 - s4) / nums;
  ans += totInv;
  printf("%0.9f", ans);
  return 0;
}