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;
int n;
int A[100050], T[400005];
long long sum[400005];
pair<int, long long> p;
double f, g, m, x, y;
void update(int x, int l, int r, int pos, long long val) {
  if (l == r) {
    T[x]++;
    sum[x] = val;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(x * 2, l, mid, pos, val);
  else
    update(x * 2 + 1, mid + 1, r, pos, val);
  T[x] = T[x * 2] + T[x * 2 + 1];
  sum[x] = sum[x * 2] + sum[x * 2 + 1];
}
pair<int, long long> query(int x, int l, int r, int pos) {
  if (r <= pos)
    return make_pair(T[x], sum[x]);
  else if (l > pos)
    return make_pair(0, 0ll);
  int mid = (l + r) / 2;
  pair<int, long long> a = query(x * 2, l, mid, pos);
  pair<int, long long> b = query(x * 2 + 1, mid + 1, r, pos);
  return make_pair(a.first + b.first, a.second + b.second);
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) scanf("%d", &A[i]);
  m = (double)n * (double)(n + 1) / 2.0;
  for (int i = (n - 1); i >= (0); i--) {
    p = query(1, 1, n, A[i]);
    x = (double)n * (double)(p.first) - (double)p.second;
    x *= (double)(i + 1);
    y += (double)(p.first) * m;
    y -= x;
    update(1, 1, n, A[i], (long long)(i));
  }
  f = 0.0;
  g = 0.5;
  for (int i = (0); i < (n); i++) {
    y += (long long)(n - i) * f;
    f += g;
    g += 0.5;
  }
  printf("%.18lf\n", y / m);
  return 0;
}