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 MAXN = 1e5 + 10, MOD = 777777777;
using ld = long double;
struct FT {
  vector<long long> s;
  FT(int n) : s(n) {}
  void upd(int pos, long long dif) {
    for (; pos < int(s.size()); pos |= pos + 1) s[pos] += dif;
  }
  long long qry(int pos) {
    long long res = 0;
    for (++pos; pos > 0; pos &= pos - 1) res += s[pos - 1];
    return res;
  }
};
int n, a[MAXN];
using ll = long double;
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], --a[i];
  FT inv(n), seg(n);
  ll no_range = 0, num_ranges = ll(n) * (n + 1) / 2;
  for (int i = 0; i < n; i++) {
    ll cur_inv = inv.qry(n - 1) - inv.qry(a[i]);
    ll tot_left = seg.qry(n - 1) - seg.qry(a[i]);
    int right = n - i;
    ll sum_range = tot_left * right;
    no_range += num_ranges * cur_inv - sum_range;
    inv.upd(a[i], 1), seg.upd(a[i], i + 1);
  }
  ld ans = ld(no_range);
  for (int range_size = 1; range_size <= n; range_size++) {
    int cnt_ranges = n - range_size + 1;
    ll exp_inv = ll(range_size) * (range_size - 1) / 2;
    ld cur = ld(cnt_ranges) * ld(exp_inv) / ld(2);
    ans += cur;
  }
  cout << fixed << setprecision(15);
  cout << ans / ld(num_ranges) << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = 1;
  while (T--) solve();
}