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;
inline char nchar() {
  static const int bufl = 1 << 20;
  static char buf[bufl], *a = NULL, *b = NULL;
  return a == b && (b = (a = buf) + fread(buf, 1, bufl, stdin), a == b) ? EOF
                                                                        : *a++;
}
inline int read() {
  int x = 0, f = 1;
  char c = nchar();
  for (; !isdigit(c); c = nchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = nchar()) x = x * 10 + c - '0';
  return x * f;
}
const int maxn = 1e5 + 1;
int n, a[maxn];
struct bit {
  long long c[maxn];
  inline int lowbit(int x) { return x & -x; }
  void inc(int x, long long d) {
    for (; x <= n; x += lowbit(x)) c[x] += d;
  }
  long long query(int x) {
    long long ret = 0;
    for (; x; x -= lowbit(x)) ret += c[x];
    return ret;
  }
} b, c;
long double get(long long n) {
  long long u = 0;
  for (long long i = 1; i <= n; ++i) u += i * (i - 1) * (n - i + 1);
  return (long double)u / 4;
}
int main() {
  n = read(), generate(a + 1, a + n + 1, read);
  long long sum = 0, s = 0, ni = 0;
  for (int i = n; i; --i) {
    s += b.query(a[i]);
    ni += c.query(a[i]);
    sum += s;
    b.inc(a[i], n - i + 1);
    c.inc(a[i], 1);
  }
  long double u = get(n) - sum;
  long double ans = u / ((long long)n * (n + 1) / 2) + ni;
  printf("%.12lf\n", (double)ans);
  return 0;
}