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 + 100;
int n, a[N];
long long s[2][N];
double Ans;
void Add(int tp, int x, long long sum) {
  while (x <= n) s[tp][x] += sum, x += (x & (-x));
}
long long Query(int tp, int x) {
  long long ret = 0;
  while (x) ret += s[tp][x], x -= (x & (-x));
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    Ans += 1.0 * Query(0, n - a[i] + 1) / (1.0 * n * (n + 1)) * i;
    Ans += 1.0 * i * (n - i + 1) * (n - i) / (2.0 * n * (n + 1));
    Add(0, n - a[i] + 1, 2 * i);
  }
  for (int i = 1; i <= n; ++i) {
    Ans += 1.0 * Query(1, n - a[i] + 1) / (1.0 * n);
    Add(1, n - a[i] + 1, n - 2 * i);
  }
  printf("%.10lf", Ans);
  return 0;
}