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;
template <class T>
void Read(T &x) {
  char c;
  bool f(0);
  while (c = getchar(), c != EOF) {
    if (c == '-')
      f = 1;
    else if (c >= '0' && c <= '9') {
      x = c - '0';
      while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0';
      ungetc(c, stdin);
      if (f) x = -x;
      return;
    }
  }
}
int p[100000 + 10], n;
long long c[100000 + 10];
double ans;
inline int lowbit(int x) { return x & -x; }
inline void update(int x, int d) {
  while (x <= n) {
    c[x] += d;
    x += lowbit(x);
  }
}
long long get_sum(int x) {
  long long ret(0);
  while (x) {
    ret += c[x];
    x ^= lowbit(x);
  }
  return ret;
}
void read() {
  Read(n);
  int x, i;
  for (i = 1; i <= n; i++) {
    Read(x);
    p[x] = i;
  }
}
void solve() {
  int i;
  for (i = 1; i <= n; i++) {
    ans += 1.0 * (n - p[i] + 1) / n / (n + 1) * get_sum(p[i]);
    update(p[i], p[i]);
  }
  for (i = n; i; i--) c[i] = 0;
  for (i = n; i; i--) {
    ans -= 1.0 * (n - p[i] + 1) / n / (n + 1) * get_sum(p[i]);
    update(p[i], p[i]);
  }
  for (i = n; i; i--) c[i] = 0;
  for (i = n; i; i--) {
    ans += get_sum(p[i]);
    update(p[i], 1);
  }
}
int main() {
  read();
  solve();
  printf("%.10lf\n", ans);
}