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 INF = 1 << 30;
const int N = 1e5 + 10;
int p[N];
long long c[N], d[N];
long long A[N], B[N];
int n;
void add(long long c[], int x, int y) {
  for (; x <= n; x += x & -x) {
    c[x] += y;
  }
}
long long query(long long c[], int x) {
  long long ret = 0;
  for (; x; x -= x & -x) {
    ret += c[x];
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &p[i]);
  }
  long long sum = 0;
  double all = (double)n * (n + 1);
  double ret = 0.0;
  for (int i = 0; i < n; ++i) {
    long long A0 = query(c, p[i]);
    long long A1 = query(d, p[i]);
    add(c, p[i], 1);
    add(d, p[i], i);
    ret += (n - i) / all * (A0 + A1);
    A0 = i - A0;
    A1 = sum - A1;
    ret += A0 - (n - i) / all * (A0 + A1);
    sum += i;
  }
  printf("%.10f\n", ret);
  return 0;
}