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 int N = int(3e5), mod = int(1e9) + 7;
int n;
int a[N], c[N];
long long t[N];
long long x[N], pref[N];
long double p[N];
long long suf[N];
void upd(int v, int val) {
  v += n - 1;
  t[v] += val;
  c[v]++;
  while (v > 1) {
    v >>= 1;
    t[v] = t[v + v] + t[v + v + 1];
    c[v] = c[v + v] + c[v + v + 1];
  }
}
long long get(int l, int r) {
  l += n - 1;
  r += n - 1;
  long long res = 0;
  while (l <= r) {
    if (l & 1) res += t[l];
    if (!(r & 1)) res += t[r];
    l = (l + 1) >> 1;
    r = (r - 1) >> 1;
  }
  return res;
}
int cnt(int l, int r) {
  l += n - 1;
  r += n - 1;
  int res = 0;
  while (l <= r) {
    if (l & 1) res += c[l];
    if (!(r & 1)) res += c[r];
    l = (l + 1) >> 1;
    r = (r - 1) >> 1;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  long long all = 1ll * n * (n + 1) / 2;
  long double res = 0;
  for (int i = 1; i <= n; i++) {
    long long cur = get(1, a[i] - 1);
    long double p = (cur + 0.0) * (n - i + 1) / all;
    res += p * 0.5;
    cur = get(a[i] + 1, n);
    p = (cur + 0.0) * (n - i + 1) / all;
    res += cnt(a[i] + 1, n);
    res -= p * 0.5;
    upd(a[i], i);
  }
  printf("%.12lf", double(res));
  return 0;
}