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 bigmod(long long b, long long p) {
  if (p == 0) return 1;
  long long my = bigmod(b, p / 2);
  my *= my;
  my %= 1000000007;
  if (p & 1) my *= b, my %= 1000000007;
  return my;
}
int setb(int n, int pos) { return n = n | (1 << pos); }
int resb(int n, int pos) { return n = n & ~(1 << pos); }
bool checkb(int n, int pos) { return (bool)(n & (1 << pos)); }
long long n, ara[200005], mx;
long long tree[200005];
long double get_val(long long x) {
  long long ret = (x * (x - 1)) / 2;
  return (long double)ret / 2.0;
}
long long read(long long ind) {
  long long sum = 0;
  while (ind > 0) {
    sum += tree[ind];
    ind -= (ind & -ind);
  }
  return sum;
}
void update(long long ind, long long val) {
  while (ind <= mx) {
    tree[ind] += val;
    ind += (ind & -ind);
  }
}
long long go(long long x) {
  long long mx = read(n);
  long long mn = read(x);
  return mx - mn;
}
int main() {
  long long i;
  scanf("%lld", &n);
  mx = n;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &ara[i]);
  }
  long long act = 0;
  for (i = 1; i <= n; i++) {
    long long my = ara[i];
    act += go(my);
    update(my, 1);
  }
  memset(tree, 0, sizeof(tree));
  long long ret = 0;
  for (i = 1; i <= n; i++) {
    long long my = ara[i];
    ret += go(my) * (n - i + 1);
    update(my, i);
  }
  long double tot = 0.0;
  for (i = 1; i <= n; i++) {
    long long my = (n - i + 1);
    tot += (my * get_val(i));
  }
  long double love = tot - ret;
  love *= 2.0;
  love /= ((n * (n + 1)));
  printf("%.20f\n", (double)(love + act));
  return 0;
}