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 inv[100005], S[4 * 100005];
vector<pair<int, int> > all;
int n, tot[4 * 100005];
void modify(int pos, int val, int id = 1, int l = 0, int r = n) {
  if (r - l < 2) {
    S[id] = val;
    tot[id] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos < mid)
    modify(pos, val, 2 * id, l, mid);
  else
    modify(pos, val, 2 * id + 1, mid, r);
  S[id] = S[2 * id] + S[2 * id + 1];
  tot[id] = tot[2 * id] + tot[2 * id + 1];
}
long long sum(int b, int e, int id = 1, int l = 0, int r = n) {
  if (b >= r || e <= l) return 0;
  if (l >= b && r <= e) return S[id];
  int mid = (l + r) >> 1;
  return sum(b, e, 2 * id, l, mid) + sum(b, e, 2 * id + 1, mid, r);
}
int nowtot(int b, int e, int id = 1, int l = 0, int r = n) {
  if (b >= r || e <= l) return 0;
  if (l >= b && r <= e) return tot[id];
  int mid = (l + r) >> 1;
  return nowtot(b, e, 2 * id, l, mid) + nowtot(b, e, 2 * id + 1, mid, r);
}
int main() {
  int i, j;
  inv[0] = 0;
  for (i = 1; i < 100005; i++) inv[i] = inv[i - 1] + i;
  double res = 0;
  scanf("%d", &n);
  long long ttt = n;
  ttt *= (n + 1);
  ttt /= 2;
  for (i = 0; i < n; i++) {
    scanf("%d", &j);
    all.push_back(make_pair(j, i));
  }
  for (i = 1; i < n; i++) {
    double tmp;
    tmp = (double)(n - i) * (double)inv[i];
    tmp /= ttt;
    res += tmp;
  }
  res *= .5;
  sort(all.begin(), all.end());
  for (i = n - 1; i >= 0; i--) {
    double p = n - all[i].second;
    p *= sum(0, all[i].second);
    p /= ttt;
    p = nowtot(0, all[i].second) - p;
    res += p;
    modify(all[i].second, all[i].second + 1);
  }
  cout << fixed << setprecision(15) << res << endl;
  return 0;
}