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 int N = 1e5 + 5;
long long c[N], s[N], n = 0;
int a[N];
long double res = 0;
int lowbit(int x) { return x & (-x); }
void add(long long *d, int x, long long val) {
  for (int i = x; i <= n; i += lowbit(i)) {
    d[i] += val;
  }
}
long long qry(long long *d, int x) {
  long long res = 0;
  for (int i = x; i; i -= lowbit(i)) {
    res += d[i];
  }
  return res;
}
void cal1() {
  long double ans = 0;
  for (long long i = 1LL; i <= n; ++i) {
    ans += (long double)(1LL * i * (n - i) * (n - i + 1)) / 2.0 / n / (n + 1);
  }
  res += ans;
}
void cal2() {
  long double sum = 0;
  for (long long i = n; i >= 1LL; --i) {
    long long x = qry(c, 1LL * (a[i] - 1LL));
    sum -= (long double)(x * ((2LL * i + 2LL * n * i) - 1LL * (n * n + n))) /
           n / (n + 1);
    sum += (long double)(2LL * i) / n / (n + 1) * qry(s, 1LL * (a[i] - 1));
    add(c, a[i], 1LL);
    add(s, a[i], 1LL * i);
  }
  res += sum;
}
int main() {
  scanf("%I64d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  cal1();
  cal2();
  printf("%.20lf", (double)res);
  return 0;
}