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 long long N = 1e5 + 10;
long long n, a[N];
double ans;
struct Bit {
  long long t[N];
  long long low(long long x) { return x & (-x); }
  long long query(long long x) {
    long long tot = 0;
    for (; x; x -= low(x)) tot += t[x];
    return tot;
  }
  void add(long long x, long long val) {
    for (; x <= n; x += low(x)) t[x] += val;
  }
} t1, t2;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (register long long i = (1); i <= (n); ++i) cin >> a[i];
  for (register long long i = (n); i >= (1); --i) {
    ans += t1.query(a[i]) * (n * n + n - 2 * i * n - 2 * i);
    ans += t2.query(a[i]) * 2 * i;
    t1.add(a[i], 1);
    t2.add(a[i], i);
  }
  for (register long long i = (1); i <= (n); ++i)
    ans += (n - i + 1) * i * (n - i) / 2.0;
  double t = 1ll * n * (n + 1);
  ans /= t;
  cout << fixed << setprecision(10) << ans << '\n';
  return 0;
}