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 long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
const long long N = 100100;
long long n;
long long a[N];
long long bit[N];
long long sum(long long x) {
  long long r = 0;
  while (x > 0) {
    r += bit[x];
    x -= (x & -x);
  }
  return r;
}
void upd(long long x, long long v) {
  while (x < N) {
    bit[x] += v;
    x += (x & -x);
  }
}
double pre[N];
int32_t main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  double den = (double)1.0 * n * (n + 1) / 2.0;
  for (long long i = 1; i < N; i++) {
    pre[i] = pre[i - 1] + (double)(1.0 * i * (i - 1)) / 4.0;
  }
  double ans = 0;
  long long tot = 0;
  for (long long i = n; i >= 1; i--) {
    tot += sum(a[i]);
    ans -= tot;
    ans += pre[n - i + 1];
    upd(a[i], n - i + 1);
  }
  ans = (1.0 * ans) / (1.0 * den);
  for (long long i = 1; i < N; i++) bit[i] = 0;
  tot = 0;
  for (long long i = n; i >= 1; i--) {
    tot += sum(a[i]);
    upd(a[i], 1);
  }
  ans += tot;
  printf("%.13lf\n", ans);
}