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>
const int N = 1e5 + 2;
long long n;
int a[N];
class TTree {
  long long bit[N];

 public:
  void Update(int id, long long val) {
    for (; id > 0; id -= id & -id) bit[id] += val;
  }
  long long Query(int id) {
    long long sum = 0;
    for (; id <= n; id += id & -id) sum += bit[id];
    return sum;
  }
} tree1, tree2;
void Solve() {
  double res = 0, seg_prob = (n * (n + 1)) / 2.;
  for (long long i = 1; i <= n; ++i) {
    res += (n - i + 1) * i * (i - 1) / 4.;
  }
  res /= seg_prob;
  for (int i = 1; i <= n; ++i) {
    int num = tree1.Query(a[i]);
    long long sum = tree2.Query(a[i]);
    res += num - sum * (n - i + 1) / seg_prob;
    tree1.Update(a[i], 1);
    tree2.Update(a[i], i);
  }
  printf("%.9f", res);
}
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  Solve();
  return 0;
}