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 maxn = 100000;
int n, a[maxn];
namespace BIT {
long long T[maxn];
void init(void) { memset(T, 0, sizeof T); }
inline int lowbit(int x) { return x & -x; }
void inc(int i, long long x) {
  while (i < n) {
    T[i] += x;
    i += lowbit(i + 1);
  }
}
long long sum(int i) {
  long long res = 0;
  while (i >= 0) {
    res += T[i];
    i -= lowbit(i + 1);
  }
  return res;
}
}  // namespace BIT
long long calc_ori_inv(void) {
  using namespace BIT;
  init();
  long long cnt = 0;
  for (int i = 0; i < n; ++i) {
    cnt += sum(n - 1) - sum(a[i]);
    inc(a[i], 1);
  }
  return cnt;
}
long long calc_tot_inv(void) {
  using namespace BIT;
  init();
  long long tot = 0, now = 0;
  for (int i = 0; i < n; ++i) {
    now += sum(n - 1) - sum(a[i]);
    tot += now;
    inc(a[i], i + 1);
  }
  return tot;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
double calc_tot_exp(void) {
  long long A = (long long)n * (n - 1), B = (long long)(n + 1) * (n + 2),
            C = 48, d;
  d = gcd(A, C);
  A /= d, C /= d;
  d = gcd(B, C);
  B /= d, C /= d;
  return (double)(A * B) / C;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]), --a[i];
  }
  long long ori_inv = calc_ori_inv();
  long long tot_inv = calc_tot_inv();
  double tot_exp = calc_tot_exp();
  double ans = (tot_exp - tot_inv) / ((long long)n * (n + 1) / 2) + ori_inv;
  printf("%.10f\n", ans);
}