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 maxn = 2e5 + 100;
const int mxlog = 20;
const long long mod = 1e9 + 7;
long long fen[maxn], a[maxn];
long long pSum(int x) {
  long long res = 0;
  for (x++; x > 0; x -= x & -x) res += fen[x];
  return res;
}
void pAdd(int x, long long val) {
  for (x++; x < maxn; x += x & -x) fen[x] += val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long kol = 0;
  for (long long i = 0; i < n; i++) {
    kol += (i - pSum(a[i]));
    pAdd(a[i], 1);
  }
  fill(fen, fen + maxn, 0);
  long double A = 0;
  for (long long i = 0; i < n; i++)
    A = A + ((i + 1) * i / (long double)4 * (n - i));
  long long inv = 0;
  for (long long i = n - 1; i >= 0; i--) {
    inv -= pSum(a[i] - 1) * (i + 1);
    pAdd(a[i], n - i);
  }
  cout << setprecision(12) << fixed
       << (kol + ((A + inv) / ((long long)n * (n + 1) / 2))) << endl;
  return 0;
}