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 n;
long long N[100006];
long long sum[1000006], S[1000006];
long long t[1000006];
long long t1[1000006];
long long lowbit(long long u) { return u & (-u); }
void add(long long u, long long w) {
  while (u <= n) {
    t[u] += w;
    u += lowbit(u);
  }
}
long long Find(long long u) {
  long long ans = 0;
  while (u) {
    ans += t[u];
    u -= lowbit(u);
  }
  return ans;
}
void add1(long long u, long long w) {
  while (u <= n) {
    t1[u] += w;
    u += lowbit(u);
  }
}
long long Find1(long long u) {
  long long ans = 0;
  while (u) {
    ans += t1[u];
    u -= lowbit(u);
  }
  return ans;
}
double ans;
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &N[i]);
  for (long long i = 1; i <= n; i++) {
    sum[i] += Find(n) - Find(N[i]);
    S[i] += Find1(n) - Find1(N[i]);
    if (sum[i] != 0)
      ans += ((double)(n * (n + 1ll) / 2ll) * sum[i] -
              (double)(n - i + 1ll) * S[i]) /
             (double)(n * (n + 1ll) / 2ll);
    add(N[i], 1);
    add1(N[i], i);
  }
  for (long long i = 2; i <= n; i++)
    ans += (double)((i - 1) * i) / 2.00 / 2.00 * (double)(n - i + 1) /
           (double)(n * (n + 1ll) / 2ll);
  printf("%.12lf\n", ans);
  return 0;
}