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>
long double S, ans;
int n, p[1000101];
struct ta {
  long long t[1000101];
  inline void Rst() { memset(t, 0, sizeof(t)); }
  inline void add(int ps, long long x) {
    while (ps <= n) {
      t[ps] += x;
      ps += ps & (-ps);
    }
  }
  inline long long query(int ps) {
    register long long ans = 0;
    while (ps) {
      ans += t[ps], ps -= ps & (-ps);
    }
    return ans;
  }
  inline long long Qc(int l, int r) {
    if (l > r) return 0;
    return query(r) - query(l - 1);
  }
} t[3];
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  S = 1.0 * n * (n + 1) / 2.0;
  for (register int i = 1; i <= n; ++i)
    ans += 1.0 * (n - i + 1) / S * 1.0 * i * (i - 1) / 4;
  t[0].Rst();
  for (register int i = 1; i <= n; ++i) {
    ans += 1.0 / S * t[0].Qc(p[i] + 1, n) * (n - i) * (n - i + 1) / 2.0;
    t[0].add(p[i], 1);
  }
  t[0].Rst();
  for (register int i = n; i >= 1; --i) {
    ans += 1.0 / S * t[0].Qc(1, p[i] - 1) * (i - 1) * i / 2.0;
    t[0].add(p[i], 1);
  }
  t[0].Rst(), t[1].Rst();
  for (register int i = 1; i <= n; ++i) {
    ans += 1.0 / S * (1.0 * i * t[0].Qc(p[i] + 1, n) - t[1].Qc(p[i] + 1, n));
    t[0].add(p[i], i);
    t[1].add(p[i], 1LL * i * i);
  }
  t[0].Rst(), t[1].Rst();
  for (register int i = n; i >= 1; --i) {
    ans +=
        1.0 / S * (1.0 * t[1].Qc(1, p[i] - 1) - 1.0 * i * t[0].Qc(1, p[i] - 1));
    t[0].add(p[i], n - i + 1);
    t[1].add(p[i], 1LL * i * (n - i + 1));
  }
  t[0].Rst(), t[1].Rst(), t[2].Rst();
  for (register int i = 1; i <= n; ++i) {
    ans += 1.0 / (2.0 * S) *
           (1.0 * (1.0 * i * i - i) * t[0].Qc(p[i] + 1, n) -
            2.0 * i * t[1].Qc(p[i] + 1, n) + 1.0 * t[2].Qc(p[i] + 1, n));
    t[0].add(p[i], 1), t[1].add(p[i], i), t[2].add(p[i], 1LL * i * i + i);
  }
  std::cout << std::fixed << std::setprecision(12) << ans << '\n';
}