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;
inline int read() {
  int sum = 0;
  char c = getchar();
  bool f = 0;
  while (c < '0' || c > '9') {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  if (f) return -sum;
  return sum;
}
const int N = 100005;
int m, a[N];
long double n;
long double c1[N], c2[N], ans;
inline void up(long double *c, int x, int v) {
  for (int i = x; i <= m; i += i & -i) c[i] += v;
}
inline long double ask(long double *c, int x) {
  long double r = 0;
  for (int i = x; i; i -= i & -i) r += c[i];
  return r;
}
int main() {
  int i, j;
  m = n = read();
  for (i = 1; i <= m; i++) a[i] = read();
  for (j = 1; j <= m; j++) ans += (n - j + 1) * j * (j - 1) / 2;
  for (i = m; i >= 1; i--) {
    ans += ask(c1, a[i] - 1) * (n * (n + 1) - (n + 1) * i * 2);
    ans += ask(c2, a[i] - 1) * i * 2;
    up(c1, a[i], 1);
    up(c2, a[i], i);
  }
  printf("%.20lf", (double)(ans / (n * (n + 1))));
  return 0;
}