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 long long N = 100010;
long long n, a[N];
double bit[N];
inline void add(long long x, double val) {
  for (; x <= n; x += x & -x) bit[x] += val;
}
inline double ask(long long x) {
  double sum = 0;
  for (; x; x -= x & -x) sum += bit[x];
  return sum;
}
int main() {
  scanf("%lld", &n);
  double tt = 0.0;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", a + i);
    tt += (ask(n) - ask(a[i]));
    add(a[i], 1.0);
  }
  double ans = 0.0;
  for (long long i = 1; i <= n; i++) {
    ans += (0.25 * i * (i - 1) + tt) * (n - i + 1);
  }
  memset(bit, 0, sizeof(bit));
  double cur = 0.0;
  for (long long i = 1; i <= n; i++) {
    double tmp = ask(n) - ask(a[i]) + cur;
    ans -= tmp;
    add(a[i], i);
    cur = tmp;
  }
  ans = ans / n / (n + 1) * 2.0;
  printf("%.12lf\n", ans);
}