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;
int n;
long long fen[200000];
long long fen2[200000];
void up(int v, long long val) {
  while (v <= n) {
    fen[v] += val;
    v = (v | (v + 1));
  }
}
long long get(int v) {
  long long val = 0;
  while (v > 0) {
    val += fen[v];
    v = (v & (v + 1)) - 1;
  }
  return val;
}
void up2(int v, long long val) {
  while (v <= n) {
    fen2[v] += val;
    v = (v | (v + 1));
  }
}
long long get2(int v) {
  long long val = 0;
  while (v > 0) {
    val += fen2[v];
    v = (v & (v + 1)) - 1;
  }
  return val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  cin >> n;
  long long N = n;
  long double ans = 0;
  long double sum = N * (N + 1) / 2;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    long long q = get(p);
    long long sk = get2(p);
    ans += q * (n - i) / sum / 2.;
    q = get(n) - q;
    sk = get2(n) - sk;
    ans += sk - q * (n - i) / sum / 2.;
    up(p, i + 1);
    up2(p, 1);
  }
  cout << (double)ans << endl;
  return 0;
}