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 bsize = 131072;
long long sum(long long i, vector<long long> &bit) {
  long long s = 0;
  while (i > 0) {
    s += bit[i];
    i -= i & (-i);
  }
  return s;
}
void add(long long i, long long x, vector<long long> &bit) {
  while (i <= bsize) {
    bit[i] += x;
    i += i & (-i);
  }
}
long long inv(vector<long long> &p) {
  long long res = 0;
  vector<long long> bit(132000, 0);
  for (long long i = p.size() - 1; i >= 0; i--) {
    res += sum(p[i], bit);
    add(p[i], 1, bit);
  }
  return res;
}
double exr(long long n) {
  double res = n;
  res *= (n - 1);
  return res / 4.0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  vector<long long> b(132000, 0);
  long long n;
  cin >> n;
  vector<long long> p(n);
  for (auto &nx : p) {
    cin >> nx;
  }
  long long base = inv(p);
  long long pat = (n * (n + 1)) / 2;
  long long tis = 0;
  double res = base, rnd = 0.0;
  res *= pat;
  for (long long i = n - 1; i >= 0; i--) {
    rnd += exr(n - i);
    tis += sum(p[i], b);
    res -= tis;
    res += rnd;
    add(p[i], n - i, b);
  }
  res /= pat;
  printf("%.12lf\n", res);
  return 0;
}