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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = ((ans % m) * (a % m)) % m;
    b /= 2;
    a = ((a % m) * (a % m)) % m;
  }
  return ans % m;
}
const long double eps = 1e-10L;
long long bit[100005];
long long a[100005];
long long n;
long long ans[100005];
void update(long long x, long long val) {
  for (long long i = x; i < 100005; i += (i & -i)) bit[i] += val;
}
long long query(long long x) {
  long long ans = 0;
  for (long long i = x; i; i -= (i & -i)) ans += bit[i];
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  double sub = 0.0;
  for (long long i = n; i >= 1; i--) {
    long long q = query(a[i]);
    ans[i] = ans[i + 1] + q;
    update(a[i], n - i + 1);
    long long diff = (n - i);
    sub += (((diff * (diff + 1) * (diff + 2))) / (double)12) - ans[i];
  }
  memset(bit, 0, sizeof bit);
  long long full = 0;
  for (long long i = n; i >= 1; i--) {
    full += query(a[i]);
    update(a[i], 1);
  }
  sub /= (double)(n * (n + 1)) / 2;
  sub += full;
  cout << fixed << setprecision(12) << sub;
}