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>
template <class T>
void Swap(T& a, T& b) {
  T c = a;
  a = b;
  b = c;
}
template <class T>
T max(T& a, T& b) {
  return a < b ? b : a;
}
template <class T>
T min(T& a, T& b) {
  return a > b ? b : a;
}
using namespace std;
double ans1, ans2, tmp, S;
int a[100010], n;
long long g[100010];
void add(int x, int y) {
  for (; x <= n; x += x & -x) g[x] += y;
}
long long query(int x) {
  long long y = 0;
  for (; x; x -= x & -x) y += g[x];
  return y;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = n; i >= 1; i--) {
    ans1 += (double)query(a[i]);
    add(a[i], 1);
  }
  memset(g, 0, sizeof g);
  for (int i = n; i >= 1; i--) {
    ans2 -= (double)query(a[i]) * i;
    add(a[i], n - i + 1);
  }
  for (int i = 1; i <= n; i++) {
    tmp += (double)(i - 1) / 2.0;
    ans2 += (double)tmp * (n - i + 1);
  }
  printf("%.10f", ans1 + 2.0 * ans2 / n / (n + 1));
  return 0;
}