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;
typedef pair<long long, long long> ll;
typedef vector<long long> vl;
typedef vector<ll> vll;
typedef vector<vl> vvl;
template <typename T>
ostream &operator<<(ostream &o, vector<T> v) {
  if (v.size() > 0) o << v[0];
  for (unsigned i = 1; i < v.size(); i++) o << " " << v[i];
  return o << endl;
}
template <typename U, typename V>
ostream &operator<<(ostream &o, pair<U, V> p) {
  return o << "(" << p.first << ", " << p.second << ") ";
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (unsigned i = 0; i < v.size(); i++) in >> v[i];
  return in;
}
template <typename T>
istream &operator>>(istream &in, pair<T, T> &p) {
  in >> p.first;
  in >> p.second;
  return in;
}
vl a(100100), seg_tree(4 * 100100, 0);
void update(long long node, long long start, long long end, long long l,
            long long r, long long val) {
  if (start > end or start > r or end < l) return;
  if (start >= l and end <= r) {
    seg_tree[node] += (end - start + 1) * val;
    return;
  }
  long long mid = (start + end) / 2;
  update(node * 2 + 1, start, mid, l, r, val);
  update(node * 2 + 2, mid + 1, end, l, r, val);
  seg_tree[node] = seg_tree[node * 2 + 1] + seg_tree[node * 2 + 2];
}
long long queryRange(long long node, long long start, long long end,
                     long long l, long long r) {
  if (start > end or start > r or end < l) return 0;
  if (start >= l and end <= r) return seg_tree[node];
  long long mid = (start + end) / 2;
  long long p1 = queryRange(node * 2 + 1, start, mid, l, r);
  long long p2 = queryRange(node * 2 + 2, mid + 1, end, l, r);
  return (p1 + p2);
}
int main() {
  std::ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  vl a(n);
  cin >> a;
  long double ans = 0.0;
  long long prev = 0;
  for (long long i = (n - 1); i >= (0); i--) {
    ans += queryRange(0, 0, 100100, 0, a[i]);
    update(0, 0, 100100, a[i], a[i], 1);
  }
  long long inv = ans;
  ans = 0.0;
  for (long long i = (1); i < (long long)n + 1; i++) {
    ans += (i * (i - 1) * (n - i + 1)) / 4.0;
    ans += inv * (n - i + 1);
  }
  fill((seg_tree).begin(), (seg_tree).end(), 0);
  for (long long i = (n - 1); i >= (0); i--) {
    ans -= (i + 1) * queryRange(0, 0, 100100, 0, a[i]);
    update(0, 0, 100100, a[i], a[i], n - i);
  }
  ans /= ((n * (n + 1)) / 2.0);
  cout.precision(15);
  cout << fixed << ans;
  return 0;
}