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 MAX = 1e5 + 5;
int arr[MAX];
int bit1[MAX];
long long bit2[MAX];
void upd1(int x) {
  while (x) {
    bit1[x] += 1;
    x -= (x & (-x));
  }
}
void upd2(int x, long long val) {
  while (x) {
    bit2[x] += val;
    x -= (x & (-x));
  }
}
long long query2(int x) {
  long long ans = 0;
  while (x < MAX) {
    ans += bit2[x];
    x += (x & (-x));
  }
  return ans;
}
int query1(int x) {
  int ans = 0;
  while (x < MAX) {
    ans += bit1[x];
    x += (x & (-x));
  }
  return ans;
}
int main() {
  long long n;
  scanf("%lld", &n);
  double ans = 0;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    long long right = (n - i + 1);
    ans += ((i - 1) * 1ll * (i)*right) / (double)(n * (n + 1) * 2);
    long long cnt = query1(a);
    long long val = query2(a);
    ans += (cnt - (((right * val / (double)(n)) * 2) / (double)(n + 1)));
    upd1(a);
    upd2(a, i);
  }
  printf("%0.9lf\n", ans);
}