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 AIB[202020][4], N;
inline int zeros(int x) { return x & (-x); }
inline void add(long long x, long long q, int a) {
  for (int i = x; i <= N; i += zeros(i)) AIB[i][a] += q;
}
inline long long comp(int x, int a) {
  long long ret = 0;
  for (int i = x; i > 0; i -= zeros(i)) ret += AIB[i][a];
  return ret;
}
int v[202020];
long long tot;
long long alt;
long long all;
int main() {
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    cin >> v[i];
  }
  for (long long i = 1; i <= N; ++i) {
    tot += comp(N, 1) - comp(v[i], 1);
    add(v[i], 1, 1);
    alt += (comp(N, 2) - comp(v[i], 2)) * (N - i + 1);
    add(v[i], i, 2);
    all += i * (i - 1) * (N - i + 1);
  }
  double ret = 1.0 * tot -
               (1.0 * alt - 1.0 * all / 4.0) / (1.0 * N * (1.0 * N + 1) / 2.0);
  printf("%.12lf\n", ret);
}