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;
long long tree[4 * 100005];
long long n, a[100005];
long long q(long long cn, long long b, long long e, long long i, long long j) {
  if (b > j or e < i) return 0;
  if (i <= b and j >= e) {
    return tree[cn];
  }
  long long lc = 2 * cn, rc = lc + 1, m = (b + e) / 2;
  return q(lc, b, m, i, j) + q(rc, m + 1, e, i, j);
}
void upd(long long cn, long long b, long long e, long long i, long long val) {
  if (b > i or e < i) return;
  if (i == b and i == e) {
    tree[cn] = val;
    return;
  }
  long long lc = 2 * cn, rc = lc + 1, m = (b + e) / 2;
  upd(lc, b, m, i, val);
  upd(rc, m + 1, e, i, val);
  tree[cn] = tree[lc] + tree[rc];
}
int main() {
  scanf("%lld", &n);
  for (long long i = (1); i <= (n); i++) scanf("%lld", &a[i]);
  long long inv = 0;
  for (long long i = (1); i <= (n); i++) {
    inv += q(1, 1, n, a[i] + 1, n);
    upd(1, 1, n, a[i], 1);
  }
  memset((tree), 0, sizeof(tree));
  long long L = 0;
  for (long long i = (1); i <= (n - 1); i++)
    L += (long long)(n - i) * (i * (i + 1)) / 2;
  long long L2 = 0;
  for (long long i = (1); i <= (n); i++) {
    L2 += (long long)(q(1, 1, n, a[i] + 1, n)) * (n - i + 1);
    upd(1, 1, n, a[i], i);
  }
  double ans = 1.0 * inv - 2.0 * (L2 * 1.0 - L / 2.0) / (1.0 * n * (n + 1));
  printf("%.15f\n", ans);
  return 0;
}