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 N = 1e5 + 100;
long long n, fen[N], fen2[N];
long long pre[N], a[N], nex[N], sum[N], b[N], bein[N], psb[N];
long double ans[N], solve[N], k2[N], k3;
void add(long long x, long long val) {
  for (; x <= n; x += x & -x) fen[x] += val;
}
void add2(long long x, long long val) {
  for (; x <= n; x += x & -x) fen2[x] += val;
}
long long get(long long x) {
  long long ans = 0;
  for (; x > 0; x -= x & -x) ans += fen[x];
  return ans;
}
long long get2(long long x) {
  long long ans = 0;
  for (; x > 0; x -= x & -x) ans += fen2[x];
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = n - a[i] + 1;
  }
  pre[0] = 0;
  add(a[0], 1);
  for (long long i = 1; i < n; i++) {
    pre[i] = pre[i - 1] + get(a[i] - 1);
    nex[i] = get(a[i] - 1);
    b[i] = nex[i];
    add(a[i], 1);
  }
  memset(fen, 0, sizeof(fen));
  add(n - a[n - 1] + 1, 1);
  for (long long i = n - 2; i >= 0; i--) {
    b[i] -= get(n - a[i]);
    bein[i] = bein[i + 1] + get(n - a[i]);
    add(n - a[i] + 1, 1);
  }
  for (long long i = n - 1; i >= 0; i--) {
    bein[i] += bein[i + 1];
    nex[i] += nex[i + 1];
    b[i] += b[i + 1];
    psb[i] = psb[i + 1] + b[i];
    sum[i] = sum[i + 1] + nex[i];
  }
  solve[1] = 0;
  for (long long i = 2; i <= n; i++) {
    solve[i] = solve[i - 1];
    solve[i] += (long double)(i * (i - 1)) / 4.0;
  }
  for (long long i = 0; i < n; i++) {
    if (i != 0) ans[i] += pre[i - 1];
    ans[i] += (long double)sum[i + 1] / (long double)(n - i);
    ans[i] += (long double)solve[n - i] / (long double)(n - i);
    ans[i] += (long double)b[i];
  }
  long double k = 0;
  for (long long i = 0; i < n; i++) {
    k += ((long double)(2 * n - 2 * i) / (long double)(n * (n + 1))) * ans[i];
  }
  memset(fen, 0, sizeof fen);
  for (long long i = 0; i < n; i++) {
    add(a[i], i);
    add2(a[i], 1);
  }
  k2[n] = 0;
  for (long long i = n - 1; i >= 0; i--) {
    k2[i] = k2[i + 1];
    if (i != n - 1) k2[i] -= (long double)b[i] - get2(a[i] - 1);
    add(a[i], -i);
    add2(a[i], -1);
    k2[i] -= (long double)get(a[i] - 1);
    k2[i] += (long double)get2(a[i] - 1) * (long double)(i - 1);
    k3 += k2[i];
  }
  k3 /= (long double)n * (n + 1) / 2.0;
  cout << fixed << setprecision(20);
  cout << k - k3 << "\n";
  return 0;
}