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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
const int mod = 1e9 + 7;
int pwr(int a, int b) {
  if (b == -1) return pwr(a, mod - 2);
  int ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * 1LL * a) % mod;
    }
    a = (a * 1LL * a) % mod;
    b >>= 1;
  }
  return ans;
}
const int N = 1e5 + 5;
pair<long long, int> bit[N];
int a[N];
void upd(int idx, pair<long long, int> p) {
  while (idx < N) {
    bit[idx].first += p.first;
    bit[idx].second += p.second;
    idx += (idx & (-idx));
  }
}
pair<long long, int> sum(int idx) {
  pair<long long, int> s = {0, 0};
  while (idx > 0) {
    s.first += bit[idx].first;
    s.second += bit[idx].second;
    idx -= idx & (-idx);
  }
  return s;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long tot = 0;
  long double ans = 0;
  long double one = 1;
  for (int i = 1; i <= n; i++) {
    pair<long long, int> p1 = sum(a[i] - 1);
    pair<long long, int> p2 = {tot - p1.first, i - 1 - p1.second};
    ans += ((n - i + 1) * one * p1.first) / (n * one * (n + 1));
    ans += p2.second - ((n - i + 1) * one * p2.first) / (n * one * (n + 1));
    upd(a[i], {i, 1});
    tot += i;
  }
  cout << setprecision(10) << fixed << ans;
  return 0;
}