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 a[100100], b[100100];
long long val[100100];
long double dp[100100];
void update(int u) {
  for (; u; u -= u & -u) b[u]++;
}
int calc(int u) {
  int ans = 0;
  for (; u < 100100; u += u & -u) ans += b[u];
  return ans;
}
void update1(int u, int Val) {
  for (; u; u -= u & -u) val[u] += Val;
}
long long calc1(int u) {
  long long ans = 0;
  for (; u < 100100; u += u & -u) ans += val[u];
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  long long cnt = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    cnt += calc(a[i]);
    update(a[i]);
  }
  for (int i = 1; i <= n; i++) dp[i] = dp[i - 1] + (long double)(i - 1) / 2;
  long double ans = 0;
  long double AA = 0;
  for (int i = 1; i <= n; i++) {
    AA += (long double)calc1(a[i]) * (n - i + 1);
    update1(a[i], i);
  }
  for (int i = 1; i <= n; i++) {
    ans += dp[i] * (n - i + 1);
  }
  ans -= AA;
  ans = ans / n / (n + 1) * 2;
  printf("%.12lf\n", (double)ans + cnt);
  return 0;
}