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 = 100009;
int n, p[N];
inline int read() {
  char c = getchar();
  int ret = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    ret = ret * 10 + c - '0';
    c = getchar();
  }
  return ret * f;
}
class Fenwick_Tree {
  double sum[N];

 public:
  inline void init() { memset(sum, 0, sizeof(sum)); }
  inline void modify(int p, int d = 1) {
    for (int i = p; i <= n; i += ((i) & -(i))) {
      sum[i] += d;
    }
  }
  inline double query(int p) {
    double ret = 0;
    for (int i = p; i; i -= ((i) & -(i))) {
      ret += sum[i];
    }
    return ret;
  }
} BIT;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    p[i] = read();
  }
  double ans = 0, psm = 0;
  for (int i = n; i; i--) {
    ans += BIT.query(p[i]);
    BIT.modify(p[i]);
  }
  ans *= n * (n + 1ll);
  BIT.init();
  for (int i = 1; i <= n; i++) {
    long long t1 = BIT.query(p[i]);
    long long t2 = i * (i - 1ll) / 2 - t1;
    ans += (psm += t1 - t2);
    BIT.modify(p[i], i);
  }
  printf("%.15lf\n", ans / n / (n + 1));
  return 0;
}