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 block_size = 320;
const long long mod = 1e9 + 7;
const long double eps = 1e-9;
const int inf = mod;
const double PI = atan(1) * 4;
template <typename T>
inline int sign(const T &a) {
  if (a < 0) return -1;
  if (a > 0) return 1;
  return 0;
}
template <typename T>
inline void in(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
template <int maxn, typename NUM>
struct BIT {
  NUM s[maxn];
  void init() { memset(s, 0, sizeof(s)); }
  BIT() { init(); }
  void update(int a, NUM value) {
    assert(a != 0);
    while (a <= maxn) {
      s[a] += value;
      a += (a & (-a));
    }
  }
  NUM query(int a) {
    NUM ans = 0;
    while (a) {
      ans += s[a];
      a -= (a & (-a));
    }
    return ans;
  }
  NUM query(int a, int b) {
    if (b < a) return 0;
    return query(b) - query(a - 1);
  }
};
BIT<100010, long long> co, cnt;
long long n;
int main() {
  cin >> n;
  double ans = 0;
  long double one = 1;
  for (long long b = 1; b < n + 1; b++) {
    int c;
    in(c);
    long long CNT = cnt.query(n) - cnt.query(c);
    long long COE = co.query(n) - co.query(c);
    cnt.update(c, 1);
    co.update(c, b);
    ans += one * CNT * (n + 1) * n / 2;
    ans -= one * COE * (n - b + 1);
  }
  for (long long len = 1; len < n + 1; len++) {
    long long cnt = n - len + 1;
    ans += one * cnt * (len - 1) * len / 4;
  }
  ans /= (n + 1) * n / 2;
  cout << setprecision(20) << ans;
  return 0;
}