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 MAX_N = 100005;
int N;
class Fenwick {
 public:
  long long sum[MAX_N];
  void update(int x, int v) {
    for (; x <= N; x += x & -x) sum[x] += v;
  }
  long long ask(int x) {
    long long res = 0;
    for (; x; x -= x & -x) res += sum[x];
    return res;
  }
} t1, t2;
int a[MAX_N];
double f[MAX_N];
int main() {
  scanf("%d", &N);
  double res1 = 0, res2 = 0;
  for (int i = 1; i <= N; ++i) scanf("%d", &a[i]);
  for (int i = N; i; --i) {
    res1 += t1.ask(a[i]);
    res2 += 1LL * t2.ask(a[i]) * i;
    t1.update(a[i], 1), t2.update(a[i], N - i + 1);
  }
  res2 /= 1LL * N * (N + 1) / 2;
  for (int i = 1; i <= N; ++i) f[i] = (double)(i - 1) / 2 + f[i - 1];
  double res3 = 0;
  for (int i = 1; i <= N; ++i) res3 += f[i] * (N - i + 1);
  res3 /= 1LL * N * (N + 1) / 2;
  printf("%.10lf\n", res1 + res3 - res2);
  return 0;
}