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>
const int MaxN = 200000 + 5;
int N;
int A[MaxN];
double F[MaxN];
inline double add(double x, double y) { return x + y; }
inline double sub(double x, double y) { return x - y; }
inline double mul(double x, double y) { return x * y; }
inline double sep(double x, double y) { return x / y; }
struct BIT {
  double b[MaxN];
  inline int lowbit(int i) { return i & -i; }
  inline void update(int x, double v) {
    for (int i = x; i <= N; i += lowbit(i)) b[i] = add(b[i], v);
  }
  inline double query(int x) {
    double res = 0;
    for (int i = x; i > 0; i -= lowbit(i)) res = add(res, b[i]);
    return res;
  }
  inline double qrange(int l, int r) { return sub(query(r), query(l - 1)); }
};
BIT B1, B2;
void init() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) scanf("%d", &A[i]);
  F[1] = 0;
  for (int i = 2; i <= N; ++i) F[i] = add(F[i - 1], (i - 1) / 2.0);
}
void solve() {
  double ans = 0;
  for (int i = 1; i <= N; ++i) {
    ans = add(ans, B1.qrange(A[i], N));
    B1.update(A[i], 1);
  }
  double sum = 0;
  for (int i = 1; i <= N; ++i) {
    double k = B2.qrange(A[i], N);
    sum = add(sum, mul(k, N - i + 1));
    B2.update(A[i], i);
  }
  for (int len = 1; len <= N; ++len) sum = sub(sum, mul(F[len], N - len + 1));
  ans = sub(ans, sep(sum, mul(N, N + 1) / 2.0));
  printf("%.12lf\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}