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 long long N = 1e5 + 5;
struct BIT {
  long long N;
  vector<long long> bit;
  void init(long long n) {
    N = n;
    bit.assign(n + 1, 0);
  }
  void update(long long idx, long long val) {
    while (idx <= N) {
      bit[idx] += val;
      idx += idx & -idx;
    }
  }
  void updateMax(long long idx, long long val) {
    while (idx <= N) {
      bit[idx] = max(bit[idx], val);
      idx += idx & -idx;
    }
  }
  long long pref(long long idx) {
    long long ans = 0;
    while (idx > 0) {
      ans += bit[idx];
      idx -= idx & -idx;
    }
    return ans;
  }
  long long rsum(long long l, long long r) { return pref(r) - pref(l - 1); }
  long long prefMax(long long idx) {
    long long ans = -2e9;
    while (idx > 0) {
      ans = max(ans, bit[idx]);
      idx -= idx & -idx;
    }
    return ans;
  }
};
long long n;
long long a[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  BIT bit;
  bit.init(1e5);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  double ans = 0;
  for (long long i = n; i >= 1; i--) {
    ans += bit.pref(a[i]);
    bit.update(a[i], 1);
  }
  bit.init(1e5);
  double extra = 0;
  for (long long i = 1; i <= n; i++) {
    extra += (n - i + 1) * bit.pref(a[i]);
    bit.update(a[i], i);
    a[i] = n - a[i] + 1;
  }
  extra /= n;
  extra /= (n + 1);
  ans += extra;
  bit.init(1e5);
  extra = 0;
  for (long long i = 1; i <= n; i++) {
    extra += (n - i + 1) * bit.pref(a[i]);
    bit.update(a[i], i);
    a[i] = n - a[i] + 1;
  }
  extra /= n;
  extra /= (n + 1);
  ans -= extra;
  cout << fixed << setprecision(12) << ans;
  return 0;
}