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;
struct fenw {
  long long n;
  vector<long long> a;
  void init(long long n) {
    this->n = n;
    a.resize(n + 1);
    fill(a.begin(), a.end(), 0);
  }
  void update(long long idx, long long d) {
    while (idx <= n) {
      a[idx] += d;
      idx += idx & -idx;
    }
  }
  long long sm(long long idx) {
    long long sum = 0;
    while (idx > 0) {
      sum += a[idx];
      idx -= idx & -idx;
    }
    return sum;
  }
};
int main() {
  long long n;
  cin >> n;
  vector<long long> A(n + 1);
  for (int i = 0; i < n; i++) cin >> A[i + 1];
  long double totalsegments = n * (n + 1) / 2;
  fenw inversions, inversionsInSegment;
  inversions.init(n);
  inversionsInSegment.init(n);
  long double res = 0;
  long long totalSegmentsAfterI = 0;
  for (long long i = n; i > 0; i--) {
    res += (long double)inversions.sm(A[i]);
    long long numSpotsToLeft = i;
    res -= (inversionsInSegment.sm(A[i])) * numSpotsToLeft / totalsegments;
    inversions.update(A[i], 1);
    long long totalSegmentsBeginningAtI = n - i + 1;
    inversionsInSegment.update(A[i], totalSegmentsBeginningAtI);
    totalSegmentsAfterI += n - i + 1;
  }
  for (long long len = 1; len <= n; len++) {
    long double expectedInversionsInPermWithLengthLen = len * (len - 1) / 4.0;
    long long numberSegmentsWithLengthLen = n - len + 1;
    res += numberSegmentsWithLengthLen * expectedInversionsInPermWithLengthLen /
           totalsegments;
  }
  cout.precision(15);
  cout << res;
  return 0;
}