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 int read() {
  int ch = getchar(), f = 0;
  int x = 0;
  while (ch < '0' || ch > '9') {
    f |= ch == '-';
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f ? -x : x;
}
int n, m;
int a[200005];
long long tree[200005], sum, ans1;
double f[200005], qzh[200005], ans;
inline void add(int p, int val) {
  for (; p <= n; p += p & -p) tree[p] += val;
}
inline long long query(int p) {
  long long ret = 0;
  for (; p; p -= p & -p) ret += tree[p];
  return ret;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++)
    a[i] = read(), ans1 += i - 1 - query(a[i]), add(a[i], 1);
  memset(tree, 0, sizeof tree);
  for (int i = 1; i <= n; i++) {
    f[i] = (f[i - 1] + (double)(i - 1) * 1.0 / 2);
    qzh[i] = qzh[i - 1] + f[i];
    sum += query(n) - query(a[i]);
    add(a[i], i);
    ans -= (sum - qzh[i]) * ((double)1.0 / ((double)n * (n + 1) / 2));
  }
  printf("%.12lf\n", ans1 + ans);
}