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;
template <typename Iter>
ostream& _out(ostream& s, Iter a, Iter b) {
  for (auto it = a; it != b; it++) s << (it == a ? "" : " ") << *it;
  return s;
}
template <typename A, typename B>
ostream& operator<<(ostream& s, pair<A, B>& p) {
  return s << "(" << p.first << " , " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T>& v) {
  return _out(s, v.begin(), v.end());
}
inline long long rit() {
  int key = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') key = -1;
    c = getchar();
  }
  long long x = 0;
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * key;
}
int __ = 1;
long long bit1[100010], bit2[100010];
int a[100010];
int n;
int lowbit(int x) { return x & -x; }
void add(long long* bit, int x, int v) {
  while (x <= n) {
    bit[x] += v;
    x += lowbit(x);
  }
}
long long sum(long long* bit, int x) {
  long long ret = 0;
  while (x > 0) {
    ret += bit[x];
    x -= lowbit(x);
  }
  return ret;
}
void init() {}
void read() {}
void solve() {}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  long long M = (long long)n * (n + 1);
  double ans = 0;
  for (int i = n; i >= 1; i--) {
    long long val1 = sum(bit2, n) - sum(bit2, a[i]);
    long long val2 = sum(bit2, a[i] - 1);
    long long cnt = sum(bit1, a[i] - 1);
    ans += 1.0 * (val1 * i - val2 * i + M * cnt) / M;
    add(bit2, a[i], n - i + 1);
    add(bit1, a[i], 1);
  }
  printf("%.9lf\n", ans);
  return 0;
}