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;
long long B[100100];
int a[100100], b[100100];
void add(int u, int x) {
  for (; u < 100100; u += u & -u) b[u] += x;
}
void Add(int u, int x) {
  for (; u < 100100; u += u & -u) B[u] += x;
}
long long calc1(int u, long long ans = 0) {
  for (; u; u -= u & -u) ans += B[u];
  return ans;
}
int calc2(int u, int ans = 0) {
  for (; u; u -= u & -u) ans += b[u];
  return ans;
}
int main() {
  cout << fixed << setprecision(20);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long long ans = 0;
  long long cnt = 0;
  for (int i = 1; i <= n; i++) {
    ans += (calc1(n) - calc1(a[i])) * (n - i + 1);
    Add(a[i], i);
    cnt += (calc2(n) - calc2(a[i]));
    add(a[i], 1);
  }
  long long p = (long long)n * (n + 1) / 2;
  long double E = cnt - (long double)ans / p;
  for (int i = 1; i <= n; i++) {
    long double t = (long double)i * (i - 1) / 4 * (n - i + 1);
    E += (long double)t / p;
  }
  cout << E << endl;
  return 0;
}