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 long long mod = 1e9 + 7;
const long long mx = 1e5 + 23;
long long bit[mx][2];
long long expo_pow(long long x, long long y) {
  if (y == 0) return 1;
  y = y % (mod - 1);
  x %= mod;
  if (y == 0) y = mod - 1;
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return res;
}
void add(long long idx, long long val) {
  for (; idx < mx; idx = idx | (idx + 1)) {
    bit[idx][0]++;
    bit[idx][1] += val;
  }
}
pair<long long, long long> get(long long idx) {
  pair<long long, long long> ans = {0, 0};
  for (; idx >= 0; idx = (idx & (idx + 1)) - 1) {
    ans.first += bit[idx][0];
    ans.second += bit[idx][1];
  }
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  double inv = 0;
  double seg_inv = 0;
  double prev = 0;
  double tot = 0;
  for (long long i = 0; i < n; i++) {
    auto it1 = get(a[i]);
    auto it2 = get(mx - 2);
    pair<long long, long long> tmp = {it2.first - it1.first,
                                      it2.second - it1.second};
    inv += tmp.first;
    seg_inv += tmp.second + prev;
    prev += tmp.second;
    tot += 1.0l * (n - i) * (i + 1) * (i) / 4.0;
    add(a[i], i + 1);
  }
  double deno = n * (n + 1) / 2.0;
  cout << setprecision(12) << fixed << inv + (tot - seg_inv) / deno << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  long long i = 1;
  while (test--) {
    solve();
  }
}