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>
long long N, BIT1[100001], BIT2[100001];
double Ans;
long long idxSum1(long long k) {
  long long sum = 0;
  while (k) {
    sum += BIT1[k];
    k -= k & (-k);
  }
  return sum;
}
long long idxSum2(long long k) {
  long long sum = 0;
  while (k) {
    sum += BIT2[k];
    k -= k & (-k);
  }
  return sum;
}
void update1(long long k) {
  while (k <= N) {
    BIT1[k]++;
    k += k & (-k);
  }
}
void update2(long long k, long long p) {
  while (k <= N) {
    BIT2[k] += p;
    k += k & (-k);
  }
}
int main(void) {
  long long i, t;
  scanf("%I64d", &N);
  for (i = 1; i <= N; i++) {
    scanf("%I64d", &t);
    Ans += (double)(N + 1 - i) * i * (i - 1) / (N * (N + 1) * 2);
    Ans += i - 1 - idxSum1(t);
    update1(t);
    Ans -= (double)(i * (i - 1) / 2 - idxSum2(t)) * (N - i + 1) * 2 /
           (N * (N + 1));
    update2(t, i);
  }
  printf("%.9lf", Ans);
  return 0;
}