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 N = 200200;
const int mod = 1e9 + 7;
long long n;
long long a[N];
pair<double, pair<double, long long> > t[4 * N];
double f[4 * N];
inline pair<double, pair<double, long long> > merge(
    pair<double, pair<double, long long> > a,
    pair<double, pair<double, long long> > b) {
  return make_pair(a.first + b.first,
                   make_pair(a.second.first + b.second.first,
                             a.second.second + b.second.second));
}
inline double get2(long long v, long long tl, long long tr, long long l,
                   long long r) {
  if (l > r) return 0;
  if (tl == l && tr == r) return f[v];
  long long tm = tl + tr >> 1;
  return get2(v + v, tl, tm, l, min(r, tm)) +
         get2(v + v + 1, tm + 1, tr, max(tm + 1, l), r);
}
inline void upd2(long long v, long long tl, long long tr, long long p,
                 double x) {
  if (tl == tr) {
    f[v] = x;
    return;
  }
  long long tm = tl + tr >> 1;
  if (p <= tm)
    upd2(v + v, tl, tm, p, x);
  else
    upd2(v + v + 1, tm + 1, tr, p, x);
  f[v] = f[v + v] + f[v + v + 1];
}
inline pair<double, pair<double, long long> > get(long long v, long long tl,
                                                  long long tr, long long l,
                                                  long long r) {
  if (l > r) return make_pair(0, make_pair(0, 0));
  if (tl == l && tr == r) return t[v];
  long long tm = tl + tr >> 1;
  return merge(get(v + v, tl, tm, l, min(r, tm)),
               get(v + v + 1, tm + 1, tr, max(tm + 1, l), r));
}
inline void upd(long long v, long long tl, long long tr, long long p,
                pair<double, pair<double, long long> > x) {
  if (tl == tr) {
    t[v] = x;
    return;
  }
  long long tm = tl + tr >> 1;
  if (p <= tm)
    upd(v + v, tl, tm, p, x);
  else
    upd(v + v + 1, tm + 1, tr, p, x);
  t[v] = merge(t[v + v], t[v + v + 1]);
}
double ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (long long i = n; i >= 1; --i) {
    double c = i / 2.0;
    pair<double, pair<double, long long> > d = get(1, 1, n, 1, a[i] - 1);
    long long cnt = d.second.second;
    double allj = d.first;
    double alli = d.second.first;
    ans += ((n - i) * (n - i + 1) - (i * i + i)) / 2.0 * cnt;
    ans += i * alli;
    ans += allj;
    double s = get2(1, 1, n, 1, a[i] - 1);
    ans += s * i / 2.0;
    s = get2(1, 1, n, a[i] + 1, n);
    ans += s * i / 2.0;
    upd2(1, 1, n, a[i], n - i + 1);
    upd(1, 1, n, a[i], make_pair(0, make_pair(i, 1)));
  }
  cout.precision(15);
  cout << fixed << ans / (n * (n + 1) / 2.0);
  return 0;
}