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 nmax = 1e5 + 5;
int n, a[nmax];
long long cntInv[nmax];
double c[nmax];
class Aib {
 public:
  long long aib[nmax];
  Aib() {
    for (int i = 0; i < nmax; ++i) {
      aib[i] = 0;
    }
  }
  int lsb(int x) { return x & (-x); }
  void update(int pos, int val) {
    for (; pos <= n; pos += lsb(pos)) {
      aib[pos] += val;
    }
  }
  long long query(int pos) {
    long long sum = 0;
    for (; pos > 0; pos -= lsb(pos)) {
      sum += aib[pos];
    }
    return sum;
  }
  long long queryRange(int a, int b) { return query(b) - query(a - 1); }
};
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  Aib aib;
  Aib aibInv;
  long long totalInv = 0;
  for (int i = n; i >= 1; --i) {
    cntInv[i] = cntInv[i + 1] + aib.queryRange(1, a[i]);
    aib.update(a[i], n - i + 1);
    totalInv += aibInv.queryRange(1, a[i]);
    aibInv.update(a[i], 1);
  }
  for (int i = 1; i <= n; ++i) {
    c[i] = 1.0 * i * (i - 1) / 4.0;
    c[i] += c[i - 1];
  }
  double ans = 0.0;
  long long numitor = 1LL * n * (n + 1) / 2;
  for (int i = n; i >= 1; --i) {
    ans += ((1.0 * (n - i + 1) * totalInv - 1.0 * cntInv[i] + c[n - i + 1]) /
            (1.0 * numitor));
  }
  cout << fixed << setprecision(10) << ans << "\n";
  return 0;
}