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 mult(int x, int y) {
  long long int ans, x1 = (long long int)x, y1 = (long long int)y;
  ans = (x1 * y1) % 1000000007;
  return (int)ans;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return a * (b / gcd(a, b)); }
long long int pow1(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return ans;
}
int mina(int arr[], int n) {
  int x = arr[0], i, pos = 0;
  for (i = 1; i < n; i++) {
    if (arr[i] < x) {
      x = arr[i];
      pos = i;
    }
  }
  return x;
}
int maxa(int arr[], int n) {
  int x = arr[0], i, pos = 0;
  for (i = 1; i < n; i++) {
    if (arr[i] > x) {
      x = arr[i];
      pos = i;
    }
  }
  return x;
}
long long int a[100005], bit1[100005], bit2[100005];
void update(long long int bit[], long long int n, long long int ind,
            long long int val) {
  while (ind <= n) {
    bit[ind] += val;
    ind += (ind & (-ind));
  }
}
long long int qry(long long int bit[], long long int ind) {
  long long int sum = 0;
  while (ind > 0) {
    sum += bit[ind];
    ind -= (ind & (-ind));
  }
  return sum;
}
long long int query(long long int bit[], long long int l, long long int r) {
  return qry(bit, r) - qry(bit, l - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n, i, sum = 0;
  cin >> n;
  for (i = 1; i < n + 1; i++) cin >> a[i];
  double ans = 0;
  double tot = 1.0 * n * (n + 1);
  tot /= (2.0);
  for (i = n + 1 - 1; i >= 1; i--) {
    double less = 1.0 * query(bit2, 1, a[i] - 1);
    double more = 1.0 * (sum - less);
    ans += 1.0 * query(bit1, 1, a[i] - 1);
    ans += (0.5 * i * (more - less) / tot);
    sum += (n - i + 1);
    update(bit1, n, a[i], 1);
    update(bit2, n, a[i], n - i + 1);
  }
  cout << fixed << setprecision(15);
  cout << ans << endl;
  return 0;
}