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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const double eps = 1e-5;
const int N = 1e5 + 10;
const int M = 4e6 + 10;
const long long MOD = 1000000007;
const int mod = 1000000007;
const int MAX = 1000000010;
const double pi = acos(-1.0);
long long g[N];
int n, a[N], f[N];
int get_k(int x) {
  int ret = 0;
  for (; x; x -= x & -x) ret += f[x];
  return ret;
}
long long get_sum(int x) {
  long long ret = 0;
  for (; x; x -= x & -x) ret += g[x];
  return ret;
}
void add_k(int x, int k) {
  for (; x <= n; x += x & -x) f[x] += k;
}
void add_sum(int x, long long sum) {
  for (; x <= n; x += x & -x) g[x] += sum;
}
int main() {
  int i, k;
  long long sumk;
  double ans = 0, fmm;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  fmm = 1.0 * n * (n + 1);
  for (i = 1; i <= n; i++) {
    k = get_k(a[i]);
    sumk = get_sum(a[i]);
    ans += 1.0 * (i - 1 - k);
    ans += (2.0 * sumk - 1.0 * i * (i - 1) / 2.0) * (n - i + 1) / fmm;
    add_k(a[i], 1);
    add_sum(a[i], i);
  }
  printf("%.16f\n", ans);
  return 0;
}