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 = 1e5 + 10;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 10;
long long a[N], cnt[N], sum[N];
void add_cnt(long long x) {
  for (; x < N; x += x & -x) cnt[x]++;
}
void add_sum(long long x, long long y) {
  for (; x < N; x += x & -x) sum[x] += y;
}
long long get_cnt(long long x) {
  long long res = 0;
  for (; x; x -= x & -x) res += cnt[x];
  return res;
}
long long get_sum(long long x) {
  long long res = 0;
  for (; x; x -= x & -x) res += sum[x];
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  double E = 0.0;
  for (long long i = 1; i <= n; i++)
    E += (i * ((n - i) * (n + 1) - (n * (n + 1) / 2 - i * (i + 1) / 2))) /
         ((n + 0.0) * (n + 1.0));
  for (long long i = n; i >= 1; i--) {
    long long c = get_cnt(a[i]), s = get_sum(a[i]);
    E += c - (i * (c * (n + 1) - s)) / (n * (n + 1) / 2.0);
    add_cnt(a[i]);
    add_sum(a[i], i);
  }
  cout << setprecision(30) << E;
  return 0;
}