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;
int mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 3e5, M = N;
vector<int> g[N];
int a[N];
long long bit[2][N];
void add(int x, int val) {
  while (x < N) bit[0][x] += val, x += x & (-x);
}
void inc(int x, int val) {
  while (x < N) bit[1][x] += val, x += x & (-x);
}
long long query(int x) {
  long long ans = 0;
  while (x) ans += bit[0][x], x -= x & (-x);
  return ans;
}
long long cnt(int x) {
  long long ans = 0;
  while (x) ans += bit[1][x], x -= x & (-x);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, n, k, j;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  double num = 0;
  long long n2 = n * 1LL * (n + 1);
  for (i = n - 1; n - 1 < -1 ? i < -1 : i > -1; n - 1 < -1 ? i += 1 : i -= 1) {
    long long x;
    long long tem = query(n) - (x = query(a[i]));
    num += ((i + 1) * tem * 1.0) / n2;
    num += cnt(a[i]) - ((i + 1) * x * 1.0) / n2;
    add(a[i], n - i);
    inc(a[i], 1);
  }
  cout << fixed << setprecision(10) << num << endl;
  return 0;
}