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;
int arr[100003];
double suff[100003], pref[100003], ppref[100003];
long long ft[100003][2];
void update(int a, int val, int tipe) {
  for (; a <= n; a += a & (-a)) ft[a][tipe] += val;
}
long long query(int a, int tipe) {
  long long res = 0;
  for (; a; a -= a & (-a)) res += ft[a][tipe];
  return res;
}
long long query(int a, int b, int tipe) {
  return query(b, tipe) - query(a - 1, tipe);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  double div = 1.0 * n * (n + 1) / 2;
  for (int i = 1; i <= n; i++) {
    pref[i] = suff[i] = 1.0 * query(arr[i], n, 0) / div;
    update(arr[i], 1, 0);
  }
  for (int i = 1; i <= n; i++) pref[i] += pref[i - 1];
  for (int i = 1; i <= n; i++) ppref[i] = pref[i] + ppref[i - 1];
  for (int i = n; i >= 1; i--) suff[i] += suff[i + 1];
  memset(ft, 0, sizeof ft);
  double cntpref = 0, cntsuff = 0, cntmid = 0, store = 0;
  for (int i = 1; i <= n; i++) {
    store += ((i)*query(arr[i], n, 0) - query(arr[i], n, 1)) / div;
    cntsuff += suff[i + 1] * i;
    cntpref += ppref[i - 1];
    cntmid += store;
    update(arr[i], 1, 0);
    update(arr[i], i, 1);
  }
  double perm = 0;
  for (int i = 1; i <= n; i++)
    perm += 1.0 * i / div * (i - 1) * (n - i + 1) / 4.0;
  double tot = cntpref + cntsuff + cntmid + perm;
  printf("%.12lf\n", tot);
}