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>
const int N = 100007;
int read() {
  int x = 0, c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  return x;
}
using db = double;
using ll = long long;
int n;
struct bit {
  ll t[N];
  void add(int p, int v) {
    for (; p; p ^= p & -p) t[p] += v;
  }
  ll ask(int p) {
    ll s = 0;
    for (; p <= n; p += p & -p) s += t[p];
    return s;
  }
} t1, t2;
int main() {
  n = read();
  db ans = 0, Ans = 0;
  for (int i = 1, a; i <= n; ++i)
    a = read(), Ans += t1.ask(a + 1),
    ans +=
        (db)(n - i + 1) * i * (i - 1) / 4.0 - (db)t2.ask(a + 1) * (n - i + 1),
    t1.add(a, 1), t2.add(a, i);
  printf("%.10lf", Ans + ans / (1ll * n * (n + 1) >> 1));
}