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, a[100005];
double ans;
struct node {
  long long t[100005];
  void add(int x, long long y) {
    while (x <= n) t[x] += y, x += x & -x;
  }
  long long query(int x) {
    long long ans = 0;
    while (x) ans += t[x], x &= x - 1;
    return ans;
  }
} t1, t2;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x + (x << 2) << 1) + (ch ^ 48);
  return x * f;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = n; i; i--)
    ans += t2.query(a[i]) - 2.0 * i * t1.query(a[i]) / (n * (n + 1.0)),
        t1.add(a[i], n - i + 1), t2.add(a[i], 1);
  for (int i = 1; i <= n; i++)
    ans += 0.5 * i * (n - i) * (n - i + 1) / (n * (n + 1.0));
  printf("%.9lf", ans);
  return 0;
}