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[100010], w[100010];
long long c[100010], tol;
long double ans;
void add(int x, int y) {
  while (x <= n) c[x] += y, x += x & (-x);
}
long long sum(int x) {
  long long t = 0;
  while (x) t += c[x], x -= x & (-x);
  ;
  return t;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    w[a[i]] = i;
  }
  for (int i = 2; i <= n; i++) tol += 1ll * i * (i - 1) * (n - i + 1) >> 1;
  for (int i = n; i; i--) {
    tol -= 1ll * (n - w[i] + 1) * sum(w[i]) << 1;
    add(w[i], w[i]);
  }
  ans = (long double)tol / ((long double)n * (n + 1));
  for (int i = 1; i <= n; i++) c[i] = 0;
  for (int i = n; i; i--) ans += sum(w[i]), add(w[i], 1);
  printf("%.20Lf", ans);
  return 0;
}