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 INF = 2000 * 1000 * 1000;
const long long LLINF = 1ll << 53;
template <class T>
void relaxmax(T &r, T v) {
  r = max(r, v);
}
template <class T>
void relaxmin(T &r, T v) {
  r = min(r, v);
}
const int MAXN = 100 * 1000;
pair<int, int> a[MAXN], b[MAXN];
pair<int, int> tmp[MAXN];
long long nbinv, snum;
int N;
void merge(pair<int, int> *p, pair<int, int> *q, int n, int m) {
  long long sj = 0;
  int i = 0, j = 0;
  int t = 0;
  while (i < n || j < m) {
    if (i == n || (j < m && p[i].first > q[j].first)) {
      sj += N - q[j].second;
      tmp[t++] = q[j++];
    } else {
      nbinv += j;
      snum += 1ll * (p[i].second + 1) * sj;
      tmp[t++] = p[i++];
    }
  }
  copy(tmp, tmp + n + m, p);
}
void merge_sort(pair<int, int> *a, int l, int r) {
  if (l >= r - 1) return;
  int m = (l + r) / 2;
  merge_sort(a, l, m);
  merge_sort(a, m, r);
  merge(a + l, a + m, m - l, r - m);
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%d", &a[i].first);
    a[i].second = i;
    b[i] = {N - a[i].first + 1, i};
  }
  merge_sort(a, 0, N);
  double ans = nbinv - 1. * snum / N / (N + 1);
  nbinv = snum = 0;
  merge_sort(b, 0, N);
  ans += 1. * snum / N / (N + 1);
  printf("%.12f\n", ans);
  return 0;
}