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 = 111111;
long long tree[maxn * 4];
void Insert(int o, int l, int r, int k, int v) {
  if (l == r) {
    tree[o] = v;
    return;
  }
  int mid = (l + r) >> 1;
  if (k <= mid)
    Insert(o * 2, l, mid, k, v);
  else
    Insert(o * 2 + 1, mid + 1, r, k, v);
  tree[o] = tree[o * 2] + tree[o * 2 + 1];
}
long long Query(int o, int l, int r, int L, int R) {
  if (L <= l && r <= R) return tree[o];
  int mid = (l + r) >> 1;
  long double ans = 0;
  if (L <= mid) ans += Query(o * 2, l, mid, L, R);
  if (R > mid) ans += Query(o * 2 + 1, mid + 1, r, L, R);
  return ans;
}
int n, x, a[maxn], f[maxn];
int main() {
  cin >> n;
  long double ans = 0;
  for (int i = 1; i <= n; i++) cin >> a[i], f[a[i]] = i;
  for (int i = 1; i <= n; i++) {
    ans += Query(1, 1, n, a[i], n);
    Insert(1, 1, n, a[i], 1);
  }
  ans *= ((long long)n * (n + 1));
  memset(tree, 0, sizeof(tree));
  long long last = 0;
  for (int i = 1; i <= n; i++) {
    ans -= (last + Query(1, 1, n, a[i], n));
    last = last + Query(1, 1, n, a[i], n);
    Insert(1, 1, n, a[i], f[a[i]] * 2);
  }
  for (int i = 1; i <= n; i++)
    ans += ((long long)(n - i + 1) * i * (i - 1) / 2);
  ans /= ((long long)n * (n + 1));
  cout << setprecision(15) << ans << endl;
}