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 nr = 1e6 + 500;
inline long long read() {
  long long ret = 0;
  long long x = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') x = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * x;
}
long long n;
long long a[nr];
inline long long lowbit(long long x) { return x & (-x); }
void MDF(long long *X, long long pos, long long add) {
  while (pos <= n) {
    X[pos] += add;
    pos += lowbit(pos);
  }
}
long long SIG(long long *X, long long pos) {
  long long ret = 0;
  while (pos) {
    ret += X[pos];
    pos -= lowbit(pos);
  }
  return ret;
}
double ans1 = 0;
double ans2 = 0;
long long BIT1[nr];
long long BIT2[nr];
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
  }
  for (int i = 1; i <= n; i++) {
    ans1 += i * (n - i) * (n - i + 1);
  }
  ans1 /= (double)(2.0 * n * (n + 1));
  for (int i = n; i >= 1; i--) {
    long long v1 = SIG(BIT1, a[i] - 1);
    ans2 += v1 * (n * n + n - 2 * n * i - 2 * i);
    long long v2 = SIG(BIT2, a[i] - 1);
    ans2 += 2 * i * v2;
    MDF(BIT1, a[i], 1);
    MDF(BIT2, a[i], i);
  }
  ans2 /= (double)(n * (n + 1));
  double ans = ans1 + ans2;
  printf("%.20lf", ans);
  return 0;
}