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 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);
template <class T, class S>
inline void add(T& a, S b) {
  a += b;
  if (a >= mod) a -= mod;
}
template <class T, class S>
inline void sub(T& a, S b) {
  a -= b;
  if (a < 0) a += mod;
}
template <class T, class S>
inline bool chkmax(T& a, S b) {
  return a < b ? a = b, true : false;
}
template <class T, class S>
inline bool chkmin(T& a, S b) {
  return a > b ? a = b, true : false;
}
struct Bit {
  long long a[N];
  void init() { memset(a, 0, sizeof(a)); }
  void modify(int x, int v) {
    for (int i = x; i < N; i += i & -i) a[i] += v;
  }
  long long sum(int x) {
    long long ans = 0;
    for (int i = x; i; i -= i & -i) ans += a[i];
    return ans;
  }
  long long query(int L, int R) {
    if (L > R) return 0;
    return sum(R) - sum(L - 1);
  }
};
int n, a[N];
Bit bit;
long long totinv;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    totinv += bit.query(a[i] + 1, n);
    bit.modify(a[i], 1);
  }
  bit.init();
  long long ret1 = 0, ret2 = 0;
  for (long long i = 1; i <= n; i++) {
    ret1 += (n - i + 1) * bit.query(a[i] + 1, n);
    bit.modify(a[i], i);
  }
  double tmp1 = 2.0 * ret1 / n / (n + 1);
  for (long long i = 1; i <= n; i++) {
    long long cnt = n - i + 1;
    ret2 += i * (i - 1) / 2 * cnt;
  }
  double tmp2 = 1.0 * ret2 / n / (n + 1);
  printf("%.12f\n", totinv - tmp1 + tmp2);
  return 0;
}