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;
int AIB[200005];
long long AIB2[200005];
int n, perm[200005];
double answer;
void update(int poz, int val) {
  for (; poz <= n; poz += ((poz ^ (poz - 1)) & poz)) AIB[poz] += val;
}
void update2(int poz, int val) {
  for (; poz <= n; poz += ((poz ^ (poz - 1)) & poz)) AIB2[poz] += val;
}
int query(int poz) {
  int s = 0;
  for (; poz; poz -= ((poz ^ (poz - 1)) & poz)) s += AIB[poz];
  return s;
}
long long query2(int poz) {
  long long s = 0;
  for (; poz; poz -= ((poz ^ (poz - 1)) & poz)) s += AIB2[poz];
  return s;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &perm[i]);
  reverse(perm + 1, perm + n + 1);
  double answer = 0;
  for (int i = 1; i <= n; i++) {
    answer += ((double)n - i + 1) * (double)i * ((double)i - 1) /
              (2 * (double)n * ((double)n + 1));
    update(perm[i], 1);
    update2(perm[i], i);
    int cnt = query(perm[i] - 1);
    long long sum = query2(perm[i] - 1);
    answer += cnt - ((double)n - i + 1) * 2 * (double)sum /
                        ((double)n * ((double)n + 1));
  }
  printf("%.9lf\n", answer);
  return 0;
}