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 long long N = 1e6 + 10;
const long long mod = 1e9 + 7;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
double ans;
struct BIT {
  long long tr[N];
  long long Bmax;
  long long lb(long long x) { return (-x) & x; }
  void init(long long n) {
    Bmax = n;
    for (long long i = 0; i <= n; i++) tr[i] = 0;
  }
  long long query(long long x) {
    if (x <= 0) return 0;
    long long res = 0;
    while (x) {
      res += tr[x];
      x -= lb(x);
    }
    return res;
  }
  void update(long long x, long long w) {
    if (x <= 0) return;
    while (x <= Bmax) {
      tr[x] += w;
      x += lb(x);
    }
  }
} t1, t2;
int a[N];
int main() {
  long long n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= n; i++) {
    ans += (double)1.00 * i * (i - 1) / 4.00 * (n - i + 1);
  }
  t1.init(n);
  t2.init(n);
  for (long long i = 1; i <= n; i++) {
    long long o1 = t1.query(n) - t1.query(a[i]);
    long long o2 = t2.query(n) - t2.query(a[i]);
    t1.update(a[i], 1);
    t2.update(a[i], i);
    ans += (double)(1.00 * n * (n + 1) * o1 * 0.5 - 1.00 * (n - i + 1) * o2);
  }
  printf("%.10lf", ans / (1.00 * n * (n + 1) / 2));
}