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 = (int)1e5 + 5;
class fenwick {
 private:
  int n;
  long long tree[N];

 public:
  fenwick(int _n) {
    n = _n;
    memset(tree, 0, sizeof(tree));
  }
  void update(int idx, long long val) {
    for (; idx; idx -= idx & -idx) tree[idx] += val;
  }
  long long read(int idx) {
    long long ans = 0;
    for (; idx <= n; idx += idx & -idx) ans += tree[idx];
    return ans;
  }
};
long long n, a[N];
int main() {
  scanf("%I64d", &n);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  double res = 0, mult = 2.0 / n / (n + 1);
  for (long long i = 1; i <= n; ++i)
    res += mult * (n - i + 1) * i * (i - 1) / 4.0;
  fenwick f1(n);
  fenwick f2(n);
  for (long long i = 1; i <= n; ++i) {
    long long cnt = f1.read(a[i]), sum = f2.read(a[i]);
    res += cnt - sum * (n - i + 1) * mult;
    f1.update(a[i], 1), f2.update(a[i], i);
  }
  printf("%.12f", res);
  return 0;
}