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;
int n;
int pos[100005];
pair<int, long long> ai[4 * 100005];
void update_ai(int node, int l, int r, int pos, pair<int, long long> val) {
  if (l == r) {
    ai[node] = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    update_ai(2 * node, l, mid, pos, val);
  else
    update_ai(2 * node + 1, mid + 1, r, pos, val);
  ai[node].first = ai[2 * node].first + ai[2 * node + 1].first;
  ai[node].second = ai[2 * node].second + ai[2 * node + 1].second;
}
void query_ai(int node, int l, int r, int lq, int rq,
              pair<int, long long> &ans) {
  if (lq <= l && r <= rq) {
    ans.first += ai[node].first;
    ans.second += ai[node].second;
    return;
  }
  int mid = (l + r) >> 1;
  if (lq <= mid) query_ai(2 * node, l, mid, lq, rq, ans);
  if (mid + 1 <= rq) query_ai(2 * node + 1, mid + 1, r, lq, rq, ans);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    pos[x] = i;
  }
  double sol = 0;
  for (int i = 1; i <= n; ++i) {
    pair<int, long long> x = make_pair(0, 0);
    query_ai(1, 1, n, 1, pos[i], x);
    sol += ((double)(n - pos[i] + 1)) / n * (((double)x.second) / (n + 1));
    x = make_pair(0, 0);
    query_ai(1, 1, n, pos[i], n, x);
    sol += x.first -
           ((double)pos[i]) / n * (x.first - ((double)x.second) / (n + 1));
    update_ai(1, 1, n, pos[i], make_pair(1, pos[i]));
  }
  printf("%.20f\n", sol);
  return 0;
}