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 int N = 1e5 + 15;
long long sum[N], a[N];
double dp[N];
int u[N];
void update(int x, int add) {
  while (x < N) sum[x] += add, x += x & (-x);
}
long long q(int x) {
  long long ret = 0;
  while (x) ret += sum[x], x = x & (x - 1);
  return ret;
}
int main() {
  int n, i, k, x, y;
  long long A = 0, B = 0;
  cin >> n;
  for (i = 1; i <= n; ++i) scanf("%d", &u[i]), update(i, 1);
  for (i = 1; i <= n; ++i) update(u[i], -1), A += q(u[i]);
  for (i = n; i >= 1; --i) update(u[i], n - i + 1);
  for (i = 1; i <= n; ++i) update(u[i], -n + i - 1), B += q(u[i]) * i;
  dp[1] = 0;
  for (i = 2; i <= n; ++i) dp[i] = dp[i - 1] + (double)(i - 1) / 2;
  double ret = A, tot = n;
  tot *= n + 1;
  tot /= 2;
  for (i = 1; i <= n; ++i) ret += (dp[i] * (n - i + 1)) / tot;
  ret -= B / tot;
  printf("%0.9lf\n", ret);
}