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>
int n, a[100003];
double ans;
int lowbit(int x) { return x & (-x); }
long long ta[100003], ta0[100003];
void add(long long *ta, int p, long long x) {
  for (int i = p; i <= n; i += lowbit(i)) ta[i] += x;
}
long long query(long long *ta, int p) {
  long long ans = 0;
  for (int i = p; i >= 1; i -= lowbit(i)) ans += ta[i];
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  long long sg = 0, all = 0, ssg = 0;
  for (int i = n; i >= 1; --i) {
    sg += query(ta, a[i] - 1);
    all += query(ta0, a[i] - 1);
    ssg += sg;
    add(ta, a[i], n - i + 1);
    add(ta0, a[i], 1);
  }
  ans = -ssg / (n * (long long)(n + 1) / 2.0);
  ans += (n * (long long)(n + 1) - 2) / 24.0;
  ans += all;
  printf("%.9f\n", ans);
  return 0;
}