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 long long md = 1e9 + 7;
long long a[100005], n;
void add(vector<long long> v[], int l, int k) {
  for (int p = 0, lol = l - 1; p < 17; lol /= 2) v[p++][lol] += k;
}
long long sum(vector<long long> v[], int l, int r) {
  long long i = 0, ans = 0;
  for (l--, r--; l < r; ++i) {
    if (l % 2) ans += v[i][l++];
    if (r % 2 == 0) ans += v[i][r--];
    if (l >= r) break;
    l /= 2, r /= 2;
  }
  if (l == r) ans += v[i][l];
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  long double ans = 0, den, ad = 0, sub = 0;
  cin >> n;
  vector<long long> invtree[17], sumtree[17];
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < 17; ++i) {
    invtree[i].resize(n / (1 << i) + 2, 0);
    sumtree[i].resize(n / (1 << i) + 2, 0);
  }
  for (int i = n; i--; add(invtree, a[i], 1)) ans += sum(invtree, 1, a[i]);
  den = 1.0 / n / (n + 1);
  for (int i = n; i--;) {
    ad += (long double)sum(sumtree, a[i], n) * (i + 1);
    sub += (long double)sum(sumtree, 1, a[i]) * (i + 1);
    add(sumtree, a[i], n - i);
  }
  ans += (ad - sub) / n / (n + 1);
  cout << fixed << setprecision(14) << ans;
}