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;
long long bit_total[100005], bit_parcial[100005];
double F[100005], sumF[100005];
int v[100005];
long long query(long long *bit, int idx) {
  long long sum = 0;
  while (idx) {
    sum += bit[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void update(long long *bit, int idx, long long delta) {
  while (idx < 100005) {
    bit[idx] += delta;
    idx += (idx & -idx);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", v + i);
  F[1] = sumF[1] = 0;
  for (int i = 2; i <= n; i++) {
    F[i] = F[i - 1] + (i - 1) / 2.0;
    sumF[i] = F[i] + sumF[i - 1];
  }
  double ans = 0;
  double deno = (1. * n * (n + 1) / 2);
  for (int i = 1; i <= n; i++) {
    int k = n - i + 1;
    ans += sumF[k] / deno;
  }
  long long total = 0;
  double parcial = 0;
  for (int i = n; i >= 1; i--) {
    total += query(bit_total, v[i]);
    update(bit_total, v[i], 1);
    parcial += i * query(bit_parcial, v[i]) / deno;
    update(bit_parcial, v[i], n - i + 1);
  }
  printf("%.20f\n", ans + (total - parcial));
  return 0;
}