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;
int inp[100005];
long long smbit[100005], cntbit[100005];
void smupd(int ind, int val) {
  while (ind < 100005) {
    smbit[ind] += val;
    ind += ind & (-ind);
  }
}
void cntupd(int ind, int val) {
  while (ind < 100005) {
    cntbit[ind] += val;
    ind += ind & (-ind);
  }
}
long long cntqry(int ind) {
  long long ans = 0;
  while (ind) {
    ans += cntbit[ind];
    ind -= ind & (-ind);
  }
  return ans;
}
long long smqry(int ind) {
  long long ans = 0;
  while (ind) {
    ans += smbit[ind];
    ind -= ind & (-ind);
  }
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &inp[i]);
  long long bada = 0;
  memset(smbit, 0, sizeof(smbit));
  memset(cntbit, 0, sizeof(cntbit));
  smupd(inp[1], 1);
  cntupd(inp[1], 1);
  long long num = 0;
  for (int i = 2; i <= n; i++) {
    bada += cntqry(n) - cntqry(inp[i]);
    long long pre = 2 * smqry(inp[i]) - smqry(n);
    pre *= n - i + 1;
    num += pre;
    smupd(inp[i], i);
    cntupd(inp[i], 1);
  }
  double ans = bada + num / (n * 1.0 * (n + 1));
  printf("%.13lf\n", ans);
}