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;
struct BIT {
  int n;
  vector<long long> bit;
  BIT(int _n) {
    n = _n;
    bit = vector<long long>(n + 1, 0);
  }
  long long sum(int i) {
    long long s = 0;
    while (i > 0) {
      s += bit[i];
      i -= i & -i;
    }
    return s;
  }
  void add(int i, long long x) {
    while (i <= n) {
      bit[i] += x;
      i += i & -i;
    }
  }
};
long long invcount(int n, const vector<int>& a) {
  long long ret = 0;
  BIT b(n);
  for (int i = n; i >= 1; --i) {
    ret += b.sum(a[i]);
    b.add(a[i], 1);
  }
  return ret;
}
int main() {
  int n;
  scanf(" %d", &n);
  vector<int> a(n + 1);
  for (int i = 1; i <= n; ++i) scanf(" %d", &a[i]);
  double ans = invcount(n, a) * 0.5 * n * (n + 1);
  for (int i = 1; i <= n; ++i) ans += 0.25 * i * (i - 1) * (n - i + 1);
  BIT bit(n);
  vector<long long> b(n + 1, 0);
  bit.add(a[n], 1);
  for (int i = n - 1; i >= 1; --i) {
    b[i] = b[i + 1] + bit.sum(a[i]);
    bit.add(a[i], n - i + 1);
  }
  for (int i = 1; i <= n; ++i) ans -= b[i];
  ans /= 0.5 * n * (n + 1);
  printf("%.12f\n", ans);
  return 0;
}