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 T>
T getint() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * p;
}
template <typename T1, typename T2>
bool umin(T1 &x, const T2 &y) {
  if (x > y) return x = y, true;
  return false;
}
template <typename T1, typename T2>
bool umax(T1 &x, const T2 &y) {
  if (x < y) return x = y, true;
  return false;
}
const int maxn = 1e6 + 10;
const int inf = (int)1e9 + 5;
const int mod = (int)1e9 + 7;
const long long llinf = (long long)1e18 + 5;
const double pi = acos(-1.0);
int n, m;
int a[maxn];
long long pref[maxn];
long long suff[maxn];
int fw[maxn];
long long fw2[maxn];
void upd(int v, int x) {
  for (; v < maxn; v |= v + 1) fw[v]++, fw2[v] += x;
}
int get(int v) {
  int ret = 0;
  for (; v >= 0; v = (v & (v + 1)) - 1) ret += fw[v];
  return ret;
}
long long get2(int v) {
  long long ret = 0;
  for (; v >= 0; v = (v & (v + 1)) - 1) ret += fw2[v];
  return ret;
}
int main() {
  int n;
  cin >> n;
  for (int i = (1); i < (n + 1); i++) cin >> a[i];
  long double ans = 0.0;
  long long ways = n * 1LL * (n + 1) / 2;
  for (int i = (1); i < (n + 1); i++) {
    long long cnt = get(n) - get(a[i]);
    long long sum = get2(n) - get2(a[i]);
    ans += cnt;
    ans -= (sum * 1LL * (n - i + 1)) * 1.0 / ways;
    upd(a[i], i);
  }
  for (int len = (1); len < (n + 1); len++) {
    ans += (n - len + 1) * 1LL * len * (len - 1) / (4.0 * ways);
  }
  cout.precision(16);
  cout << ans << endl;
  return 0;
}