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;
long long BIT[6][N];
int n, arr[N];
inline long long getPrefix(int z, int i) {
  i = n - i - 1;
  ++i;
  long long ret = 0;
  while (i) ret += BIT[z][i], i -= i & -i;
  return ret;
}
inline void add(int z, int i, long long v) {
  i = n - i - 1;
  ++i;
  while (i < N) BIT[z][i] += v, i += i & -i;
}
long long f(long long x) { return x * (x + 1) / 2; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &arr[i]), arr[i]--;
  double out = 0;
  double fn = f(n);
  for (int j = 0; j < n; j++) {
    double i = getPrefix(0, arr[j]);
    double i1 = getPrefix(1, arr[j]);
    double fi = getPrefix(2, arr[j]);
    double fni = getPrefix(3, arr[j]);
    double ci = getPrefix(4, arr[j]);
    double temp = ci * f(n - j - 1) / fn;
    temp += (fni - ci * f(n - j - 1)) / fn;
    temp += (i1 * j - fi * 2) / fn;
    temp += fi / fn;
    out += temp;
    add(0, arr[j], j);
    add(1, arr[j], j + 1);
    add(2, arr[j], f(j));
    add(3, arr[j], f(n - j - 1));
    add(4, arr[j], 1);
  }
  for (long long len = 1; len <= n; len++)
    out += len * (len - 1) / 4.0 * (n - len + 1) / fn;
  printf("%.9lf\n", out);
}