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>
const int maxn = 1e5 + 10;
using namespace std;
long long gcd(long long p, long long q) { return q == 0 ? p : gcd(q, p % q); }
long long qpow(long long p, long long q) {
  long long f = 1;
  while (q) {
    if (q & 1) f = f * p;
    p = p * p;
    q >>= 1;
  }
  return f;
}
inline void umax(int &p, int q) {
  if (p < q) p = q;
}
inline void umin(int &p, int q) {
  if (p > q) p = q;
}
inline long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, k, t;
long long a[maxn], d[maxn];
double ans, now, pre;
void add(int x, int y) {
  for (int i = x; i <= n; i += i & (-i)) d[i] += y;
}
long long get(int x) {
  long long ret = 0;
  for (int i = x; i; i -= i & (-i)) ret += d[i];
  return ret;
}
void init() {
  long long cnt = 0;
  for (int i = 1; i <= n; i++) {
    cnt += get(n) - get(a[i]);
    add(a[i], 1);
  }
  for (int i = 1; i <= n; i++)
    ans += (n - i + 1) * ((double)i * (i - 1) / 4 + cnt);
  memset(d, 0, sizeof(d));
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) a[i] = read();
  init();
  for (i = 1; i <= n; i++) {
    now = get(n) - get(a[i]) + pre;
    ans -= now;
    pre = now;
    add(a[i], i);
  }
  ans /= (double)n * (n + 1) / 2;
  printf("%.10f\n", ans);
  return 0;
}