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 = 1e5 + 5;
long long ic_bit[MAX], bad_bit[MAX];
int n;
int perm[MAX];
inline void update(long long bit[], int idx, int val) {
  while (idx <= n) {
    bit[idx] += val;
    idx += idx & -idx;
  }
}
inline long long query(long long bit[], int idx) {
  long long sum = 0;
  while (idx) {
    sum += bit[idx];
    idx -= idx & -idx;
  }
  return sum;
}
int main() {
  scanf("%d", &n);
  for (int i = int(1); i < int(n + 1); i++) {
    scanf("%d", perm + i);
  }
  long long ic = 0, bad = 0;
  for (int i = int(n); i >= int(1); i--) {
    ic += query(ic_bit, perm[i]);
    bad += query(bad_bit, perm[i]) * i;
    update(ic_bit, perm[i], 1);
    update(bad_bit, perm[i], n - i + 1);
  }
  long double expected = 0.0;
  for (int i = int(1); i < int(n + 1); i++) {
    long double cur =
        (long double)(n - i + 1) * ((long double)(1LL * i * (i - 1)) / 4.0);
    expected += cur;
  }
  long double num = expected - bad;
  long double den = (1LL * n * (n + 1)) / 2;
  cout << fixed << setprecision(15) << ic + num / den << '\n';
  return 0;
}