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 NMAX = 100000 + 5;
int n;
int v[NMAX];
long long int aibCnt[NMAX];
long long int aibSum[NMAX];
inline int lsb(int node) { return node & (-node); }
void update(long long int aib[], int where, long long int val) {
  for (; where <= n; where += lsb(where)) aib[where] += val;
}
long long int query(long long int aib[], int where) {
  long long int ans = 0;
  for (; where; where -= lsb(where)) ans += aib[where];
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> v[i];
  cout << fixed << setprecision(16);
  long double ans = 0;
  long long int sum = 0;
  for (int j = 1; j <= n; ++j) {
    ans += sum * (n - j + 1LL);
    sum += j;
  }
  for (int i = n; i; --i) {
    ans += query(aibCnt, v[i]) * (n * (n + 1LL) - 2 * i * (n + 1LL));
    ans += 2 * i * query(aibSum, v[i]);
    update(aibCnt, v[i], 1);
    update(aibSum, v[i], i);
  }
  ans = ans / (n + 1.0) / n;
  cout << ans << '\n';
  return 0;
}