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;
long long yy;
long double ans1 = 0, ans2;
int n;
long long tre1[100005], tre2[100005], inv1[100005], inv2[100005];
int c;
int lowb(int x) { return -x & x; }
long long ask1(int x) {
  long long nowans = 0;
  while (x) {
    nowans += tre1[x];
    x -= lowb(x);
  }
  return nowans;
}
long long ask2(int x) {
  long long nowans = 0;
  while (x) {
    nowans += tre2[x];
    x -= lowb(x);
  }
  return nowans;
}
void add1(int x) {
  while (x <= n) {
    tre1[x]++;
    x += lowb(x);
  }
}
void add2(int x, long long y) {
  while (x <= n) {
    tre2[x] += y;
    x += lowb(x);
  }
}
int main() {
  cin >> n;
  ans2 = 0.5 * (n + 1) * n;
  for (int i = 1; i <= n; i++) ans1 += 0.25 * (i - 1) * i * (n - i + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c);
    ans1 += ans2 * (ask1(n) - ask1(c));
    ans1 -= 1.0 * (n - i + 1) * (ask2(n) - ask2(c));
    add1(c);
    add2(c, i);
  }
  printf("%.12f\n", double(ans1 / ans2));
  return 0;
}