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;
int a[100010];
long long bit[100010];
int sz;
void init() {
  for (int i = 0; i <= sz; i++) bit[i] = 0;
}
void add(int x, int inc) {
  while (x <= sz) {
    bit[x] += inc;
    x += x & (-x);
  }
}
long long query(int x) {
  long long ret = 0;
  while (x) {
    ret += bit[x];
    x -= x & (-x);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  long long avg = 0;
  for (int i = 1; i <= n; i++) {
    avg += 1LL * i * (i - 1) / 2 * (n - i + 1);
  }
  sz = n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long inv = 0;
  for (int i = 1; i <= n; i++) {
    add(a[i], 1);
    inv += a[i] - query(a[i]);
  }
  long double ans = inv;
  reverse(a + 1, a + n + 1);
  init();
  long long ret = 0;
  for (int i = 1; i <= n; i++) {
    ret -= 1LL * query(a[i]) * (n - i + 1);
    add(a[i], i);
  }
  long double tot = (1LL * n * (n + 1) / 2);
  ans = inv + avg / tot / 2.0 + ret / tot;
  cout << fixed << setprecision(20) << ans << endl;
  return 0;
}