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;
class SegmentTree {
 public:
  vector<long long> data;
  int n;
  SegmentTree(int n) {
    this->n = n;
    data.assign(2 * n, 0);
  }
  void update(int idx, int val) {
    idx += n;
    data[idx] = val;
    while (idx > 1) {
      idx >>= 1;
      data[idx] = data[idx << 1] + data[idx << 1 | 1];
    }
  }
  long long range(int left, int right) {
    left += n;
    right += n;
    long long result = 0;
    while (left < right) {
      if (left & 1) result += data[left++];
      if (right & 1) result += data[--right];
      left >>= 1;
      right >>= 1;
    }
    return result;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    v[i]--;
  }
  double segment_cnt = (double)n * (n + 1) / 2;
  vector<int> inv_greater_before(n, 0);
  SegmentTree st(n);
  for (int i = 0; i < n; i++) {
    inv_greater_before[i] = st.range(v[i], n);
    st.update(v[i], 1);
  }
  double all_inversions =
      accumulate(inv_greater_before.begin(), inv_greater_before.end(), 0LL);
  vector<long long> s(n, 0);
  SegmentTree st2(n);
  for (int i = n - 1; i >= 0; i--) {
    if (i < n - 1) s[i] = s[i + 1];
    s[i] += st2.range(0, v[i]);
    st2.update(v[i], n - i);
  }
  double all_segments = 0;
  for (int i = 0; i < n; i++) all_segments += s[i] / segment_cnt;
  double all_random_segments = 0;
  for (int length = 1; length <= n; length++) {
    all_random_segments +=
        (double)length * (length - 1) / 4 * (n - length + 1) / segment_cnt;
  }
  cout << all_inversions + all_random_segments - all_segments << endl;
  return 0;
}