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 = 1.01e9;
const double eps = 1e-9;
const int N = 1e5 + 10;
struct fen {
  long long t[N];
  long long get(int x) {
    x += 1;
    long long res = 0;
    for (; x > 0; x -= x & -x) res += t[x];
    return res;
  }
  void upd(int x, int dx) {
    x += 1;
    for (; x < N; x += x & -x) t[x] += dx;
  }
};
fen A, B;
int a[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; (i) < (n); ++i) scanf("%d", a + i);
  double res = 0;
  double total = n * (long long)(n + 1) / 2;
  for (int i = 0; (i) < (n); ++i) {
    int x = n - a[i];
    int cnt = A.get(x);
    long long sum = B.get(x);
    double CNT = sum * (n - i);
    res += total * cnt - CNT;
    A.upd(x, 1);
    B.upd(x, i + 1);
  }
  for (int len = 1; len <= n; ++len) {
    res += (n - len + 1) * (double)len * (len - 1) / 4;
  }
  res /= total;
  printf("%.10f\n", res);
  return 0;
}