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 int Fenwick[160005];
int lsb(int x) { return x & (-x); }
void update(int pos, long long int val) {
  while (pos <= 160000) {
    Fenwick[pos] += val;
    pos += lsb(pos);
  }
}
long long int query(int pos) {
  long long int ans = 0;
  while (pos) {
    ans += Fenwick[pos];
    pos -= lsb(pos);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> arr(n);
  long long int inv = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; i++) {
    inv += i - query(arr[i]);
    update(arr[i], 1);
  }
  memset(Fenwick, 0, sizeof(Fenwick));
  double invsum = 0;
  double prevbad = 0;
  for (long long int i = 0; i < n; i++) {
    long long int quer = query(arr[i]);
    long long int bad = (i * (i + 1)) / 2 - quer;
    update(arr[i], i + 1);
    invsum += prevbad + bad;
    prevbad += bad;
  }
  double expinvsum = 0;
  for (long long int len = 1; len <= n; len++) {
    expinvsum += ((n - len + 1) * (len) * (len - 1)) / 4.0;
  }
  double ans = expinvsum - invsum;
  ans /= n;
  ans /= (n + 1);
  ans *= 2;
  cout << setprecision(15) << inv + ans;
  return 0;
}