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>
const int MAXN = 100005;
int N, a[MAXN];
long long sum1[MAXN], sumI[MAXN];
int lowbit(int x) { return x & -x; }
long long query(long long arr[], int x) {
  long long ans = 0;
  while (x) ans += arr[x], x -= lowbit(x);
  return ans;
}
void modify(long long arr[], int x, long long val) {
  while (x <= N) arr[x] += val, x += lowbit(x);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) scanf("%d", &a[i]);
  double ans = 0;
  for (long long i = 1; i < N; ++i)
    ans += (i * (N + 1 - i) * (N - i)) / 2.0 / (double)N;
  for (long long i = N; i >= 1; --i) {
    ans +=
        (query(sumI, a[i] - 1) * 2 * i +
         query(sum1, a[i] - 1) * (((long long)N) * (N + 1) - (N * 2 + 2) * i)) /
        (double)N;
    modify(sumI, a[i], i);
    modify(sum1, a[i], 1);
  }
  printf("%.15lf\n", ans / (double(N + 1)));
  return 0;
}