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;
struct BIT {
  vector<long long int> bit;
  long long int n;
  void build(long long int n) {
    this->n = n;
    bit.assign(n, 0);
  }
  long long int sum(long long int r) {
    long long int s = 0;
    while (r >= 0) {
      s += bit[r];
      r &= r + 1;
      --r;
    }
    return s;
  }
  long long int sum(long long int l, long long int r) {
    if (r < l) return 0;
    return sum(r) - sum(l - 1);
  }
  void add(long long int id, long long int delta) {
    while (id < n) {
      bit[id] += delta;
      id |= id + 1;
    }
  }
};
long long int n;
long long int a[100100];
pair<long long int, long long int> sumlr[100100];
long long int inv_tot;
int main() {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    --a[i];
  }
  reverse(a, a + n);
  BIT inv1;
  inv1.build(n);
  inv_tot = 0;
  for (long long int i = 0; i < n; i++) {
    inv1.add(a[i], 1);
    inv_tot += inv1.sum(0, a[i] - 1);
  }
  BIT inv2;
  inv2.build(n);
  sumlr[0] = {0, 0};
  inv2.add(a[0], 1);
  for (long long int i = 1; i < n; i++) {
    inv2.add(a[i], i + 1);
    sumlr[i].second = sumlr[i - 1].second + inv2.sum(0, a[i] - 1);
    sumlr[i].first = sumlr[i - 1].first + sumlr[i].second / ((n * (n + 1)) / 2);
    sumlr[i].second %= ((n * (n + 1)) / 2);
  }
  pair<long long int, long long int> final_sum = {0, 0};
  for (long long int i = 0; i < n; i++) {
    final_sum.first += sumlr[i].first;
    final_sum.second += sumlr[i].second;
    final_sum.first += final_sum.second / ((n * (n + 1)) / 2);
    final_sum.second %= ((n * (n + 1)) / 2);
  }
  long double ans =
      ((long double)inv_tot) +
      ((long double)((n - 1) * (n + 2))) / ((long double)24) -
      ((long double)final_sum.first) -
      ((long double)final_sum.second) / ((long double)((n * (n + 1)) / 2));
  printf("%.30Lf\n", ans);
  return 0;
}