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 getint() {
  bool f = 0;
  int x = 0;
  char c = getchar();
  if (c == EOF) return EOF;
  while (c < '0' || c > '9') {
    if (c == '-') f = 1;
    c = getchar();
    if (c == EOF) return EOF;
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (f) return -x;
  return x;
}
const int maxn = 100000;
int n, a[maxn + 2];
long double p;
long long c[maxn + 2], ans, f[maxn + 2], sum;
inline void add(int x, long long d) {
  while (x <= n) c[x] += d, x += x & -x;
}
inline long long ask(int x) {
  long long tot = 0;
  while (x) tot += c[x], x -= x & -x;
  return tot;
}
int main() {
  n = getint();
  for (int i = 1; i <= n; i++) {
    a[i] = getint();
    sum -= ask(a[i] - 1);
    sum += i - 1;
    add(a[i], 1);
  }
  memset(c + 1, 0, n << 3);
  for (int i = n; i >= 1; i--) {
    f[i] = ask(a[i] - 1) * (long long)i;
    add(a[i], n - i + 1);
    f[0] += f[i];
  }
  for (int i = 2; i <= n; i++) ans += (long long)i * (i - 1) * (n - i + 1);
  ans -= f[0] << 2LL;
  long long temp = n;
  temp *= n + 1;
  temp <<= 1;
  p = (long double)ans / (long double)temp;
  p += sum;
  cout << setprecision(15) << p << endl;
  return 0;
}