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;
const long long N = 100005;
long long num1[N], num2[N], x, n;
double f[N];
void insert(long long x, long long y) {
  for (long long i = x; i < N; i += i & -i) num1[i]++, num2[i] += y;
}
long long find1(long long x) {
  long long ans = 0;
  for (; x; x -= x & -x) ans += num1[x];
  return ans;
}
long long find2(long long x) {
  long long ans = 0;
  for (; x; x -= x & -x) ans += num2[x];
  return ans;
}
signed main() {
  scanf("%lld", &n);
  long long ans1 = 0, ans2 = 0;
  double s = 0;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &x);
    x = n - x + 1;
    ans1 += find1(x);
    ans2 += find2(x) * (n - i + 1);
    insert(x, i);
  }
  for (long long i = 1; i <= n; i++) f[i] = i * (i - 1) / 2 + f[i - 1];
  for (long long i = 1; i <= n; i++) s += f[i];
  printf("%.10lf\n",
         ans1 - (double)ans2 / (n * (n + 1) / 2) + (double)s / (n * (n + 1)));
}