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;
using llint = long long;
const int N = 1e5 + 10;
int n;
int a[N];
double calc(int x) { return 1.0 * x * (x - 1) / 2 / 2; }
struct Bit {
  llint u[N];
  static const int B = 5;
  void clear() { memset(u, 0, sizeof(u)); }
  void modify(int x, llint v) {
    for (x += B; x < N; x += x & -x) {
      u[x] += v;
    }
  }
  llint getsum(int x) {
    llint s = 0;
    for (x += B; x; x -= x & -x) {
      s += u[x];
    }
    return s;
  }
} bit;
double go(llint t[], llint j) {
  double a1 = 1.0 * (t[2] - t[1]) / n / (n + 1);
  double a2 = 1.0 * (j * j * t[0] - 2 * j * t[1] + t[2] - j * t[0] + t[1]) / n /
              (n + 1);
  double a3 = 1.0 * (n - j + 1) * (n - j) * t[0] / n / (n + 1);
  return a1 + a2 + a3;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  double ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += bit.getsum(n) - bit.getsum(a[i]);
    bit.modify(a[i], 1);
  }
  llint tot = 1ll * n * (n + 1) / 2;
  bit.clear();
  for (int i = 1; i <= n; ++i) {
    llint t_big = bit.getsum(n) - bit.getsum(a[i]);
    llint t_small = bit.getsum(a[i]);
    ans += -0.5 * t_big * (n - i + 1) / tot;
    ans += 0.5 * t_small * (n - i + 1) / tot;
    bit.modify(a[i], i);
  }
  printf("%.10lf\n", ans);
  return 0;
}