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;
const int MAXN = 1e5 + 5;
long long int n, inv = 0, all, seg[MAXN * 4], a[MAXN];
long double ex = 0, ex2;
long long int get(int l, int r, int b = 0, int e = n, int ind = 1) {
  if (l <= b && r >= e) return seg[ind];
  long long int ret = 0;
  int mid = (b + e) / 2;
  if (l < mid) ret += get(l, r, b, mid, ind * 2);
  if (r > mid) ret += get(l, r, mid, e, ind * 2 + 1);
  return ret;
}
void add(long long int val, int x, int b = 0, int e = n, int ind = 1) {
  seg[ind] += val;
  if (b + 1 == e) return;
  int mid = (b + e) / 2;
  if (x < mid)
    add(val, x, b, mid, ind * 2);
  else
    add(val, x, mid, e, ind * 2 + 1);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  all = ((n + 1) * n) / 2ll;
  vector<pair<int, int>> v;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (long long int i = 1; i <= n; i++) {
    long double tmp = ((long double)(n - i + 1) * i * (i - 1)) / (long double)4;
    ex += tmp;
  }
  for (int i = n - 1; i >= 0; i--) {
    ex2 += ((i + 1) * get(0, a[i]));
    add(n - i, a[i]);
  }
  for (int i = 0; i < 4 * MAXN; i++) seg[i] = 0;
  for (int i = n - 1; i >= 0; i--) {
    inv += get(0, a[i]);
    add(1, a[i]);
  }
  cout << fixed << setprecision(10);
  cout << inv + ((ex - ex2) / all) << endl;
  return 0;
}