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[200007];
struct Tree {
  long long sum[200007];
  void add(int x, long long d) {
    for (; x <= n; x += x & -x) sum[x] += d;
  }
  long long query(int x) {
    long long ans = 0;
    for (; x; x -= x & -x) ans += sum[x];
    return ans;
  }
} T1, T2;
int main() {
  double ans;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  ans = 0;
  for (int j = 1; j <= n; j++) ans += 1.0 * j * (j - 1) * (n - j + 1) / 4;
  long long cnt = 0;
  for (int i = n; i >= 1; i--) {
    int x = a[i];
    cnt += T1.query(x);
    ans -= 1.0 * i * T2.query(x);
    T1.add(x, 1), T2.add(x, n - i + 1);
  }
  ans /= 1.0 * n * (n + 1) / 2;
  ans += cnt;
  printf("%.10lf\n", ans);
  return 0;
}