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;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int n, arr[100005];
long long BIT[100005], cnt[100005];
void update(int idx, int val) {
  while (idx <= n) {
    BIT[idx] += val;
    cnt[idx]++;
    idx += idx & (-idx);
  }
}
long long query(int idx) {
  long long ans = 0;
  while (idx) {
    ans += BIT[idx];
    idx -= idx & (-idx);
  }
  return ans;
}
long long query_cnt(int idx) {
  long long ans = 0;
  while (idx) {
    ans += cnt[idx];
    idx -= idx & (-idx);
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
  double ans = 0.0;
  long long tot = 0;
  long long z = 1LL * n * (n + 1) / 2;
  for (int i = n; i >= 1; i--) {
    long long less = query(arr[i] - 1);
    long long more = tot - less;
    double temp = 0.0;
    if (less != 0) {
      temp += query_cnt(arr[i] - 1) - 0.5 * (1.0 * i) * less / z;
    }
    if (more != 0) {
      temp += 0.5 * (1.0 * i) * more / z;
    }
    ans += temp;
    tot += n - i + 1;
    update(arr[i], n - i + 1);
  }
  printf("%.10lf\n", ans);
  return 0;
}