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 long long maxn = 1e6 + 10;
struct node {
  long double inv;
  long double sum;
};
long long a[maxn];
node seg[maxn];
void add(long long l, long long r, long long ind, long long val, long long x) {
  if (l + 1 >= r) {
    seg[x].inv = val;
    return;
  }
  long long mid = (l + r) / 2;
  if (ind < mid) {
    add(l, mid, ind, val, x * 2);
  } else {
    add(mid, r, ind, val, x * 2 + 1);
  }
  seg[x].inv = seg[2 * x].inv + seg[2 * x + 1].inv;
}
void add2(long long l, long long r, long long ind, long long x) {
  if (l + 1 >= r) {
    seg[x].sum++;
    return;
  }
  long long mid = (l + r) / 2;
  if (ind < mid) {
    add2(l, mid, ind, 2 * x);
  } else {
    add2(mid, r, ind, 2 * x + 1);
  }
  seg[x].sum = seg[x * 2].sum + seg[x * 2 + 1].sum;
}
long long get(long long l, long long r, long long L, long long R, long long x) {
  if (l >= R or r <= L) {
    return 0;
  }
  if (l >= L && r <= R) {
    return seg[x].inv;
  }
  long long mid = (l + r) / 2;
  return get(l, mid, L, R, x * 2) + get(mid, r, L, R, x * 2 + 1);
}
long long get2(long long l, long long r, long long L, long long R,
               long long x) {
  if (l >= R or r <= L) {
    return 0;
  }
  if (l >= L && r <= R) {
    return seg[x].sum;
  }
  long long mid = (l + r) / 2;
  return get2(l, mid, L, R, x * 2) + get2(mid, r, L, R, x * 2 + 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  long double esum = 0;
  long double isum = 0;
  long double tsum = 0;
  for (long long i = 1; i <= n; i++) {
    esum += (long double)1 * (n - i + 1) * i * (i - 1) / (long double)4;
  }
  for (long long i = 0; i < n; i++) {
    isum += (long double)1 * (n - i) * get(0, n, a[i] + 1, n, 1);
    tsum += (long double)1 * get2(0, n, a[i] + 1, n, 1);
    add(0, n, a[i], i + 1, 1);
    add2(0, n, a[i], 1);
  }
  long double total = n * (n + 1) / 2;
  cout << fixed << setprecision(10)
       << (long double)((esum - isum) / total) + tsum << endl;
  return 0;
}