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;
using ll = long long;
using db = double;
using pii = pair<int, int>;
using vi = vector<int>;
mt19937 mrand(time(0));
ll get(ll r) { return ((ll)mrand() * mrand() % r + r) % r; }
ll get(ll l, ll r) { return get(r - l + 1) + l; }
int n, a[100100];
db ans, tr2[100100], tr1[100100];
void upd(db* tr, int x, db y) {
  for (; x <= n; x += x & -x) tr[x] += y;
}
db qry(db* tr, int x) {
  db res = 0;
  for (; x; x -= x & -x) res += tr[x];
  return res;
}
signed main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) ans += 1.0 * i * (n + 1 - i) * (n - i);
  ans /= 2.0 * n * (n + 1);
  for (int i = 1; i <= n; i++) {
    ans += qry(tr1, n - a[i] + 1) - (n - i + 1) * qry(tr2, n - a[i] + 1);
    upd(tr1, n - a[i] + 1, 1), upd(tr2, n - a[i] + 1, 2.0 * i / n / (n + 1));
  }
  printf("%.20lf\n", ans);
  fprintf(stderr, "time=%.4f\n", (db)clock() / CLOCKS_PER_SEC);
  return 0;
}