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;
using ll = long long;
using ld = long double;
const int maxn = 1 << 17;
ll n, a[maxn], f[maxn];
ld ans = 0;
void add(int i, int v) {
  for (; i <= n; i += i & -i) f[i] += v;
}
ll get(int i) {
  ll res = 0;
  for (; i; i -= i & -i) res += f[i];
  return res;
}
ll get(int l, int r) { return get(r) - (l - 1 ? get(l - 1) : 0); }
void clear() { memset(f, 0, sizeof f); }
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  ll pairs = n * 1ll * (n + 1) / 2;
  ll inversions = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    inversions += get(a[i], n);
    add(a[i], 1);
  }
  clear();
  ld exp_added = 0;
  ld exp_removed = 0;
  for (int i = 1; i <= n; i++) {
    exp_added += get(1, a[i]) * (n - i + 1) * 0.5 / pairs;
    exp_removed += get(a[i], n) * (n - i + 1) * 0.5 / pairs;
    add(a[i], i);
  }
  cout << fixed << setprecision(15) << inversions + exp_added - exp_removed
       << '\n';
}