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 n;
int a[100007];
int inv[100007];
struct node {
  double val;
  int cnt;
  node() {
    val = 0.0;
    cnt = 0;
  }
  node(double _val, int _cnt) {
    val = _val;
    cnt = _cnt;
  }
};
node unite(node p1, node p2) {
  node ret = p1;
  ret.val += p2.val;
  ret.cnt += p2.cnt;
  return ret;
}
class Tree {
 public:
  node tr[4 * 100007];
  void init(int where, int IL, int IR) {
    tr[where] = node(0.0, 0);
    if (IL == IR) {
      return;
    }
    int mid = (IL + IR) / 2;
    init(2 * where, IL, mid);
    init(2 * where + 1, mid + 1, IR);
  }
  void update(int where, int IL, int IR, int pos, double add) {
    tr[where].val += add;
    ++tr[where].cnt;
    if (IL == IR) {
      return;
    }
    int mid = (IL + IR) / 2;
    if (pos <= mid) {
      update(2 * where, IL, mid, pos, add);
    } else {
      update(2 * where + 1, mid + 1, IR, pos, add);
    }
  }
  node query(int where, int IL, int IR, int CURL, int CURR) {
    if (IL > IR || CURL > CURR) {
      return node(0.0, 0);
    }
    if (CURR < IL || IR < CURL) {
      return node(0.0, 0);
    }
    if (CURL <= IL && IR <= CURR) {
      return tr[where];
    }
    int mid = (IL + IR) / 2;
    return unite(query(2 * where, IL, mid, CURL, CURR),
                 query(2 * where + 1, mid + 1, IR, CURL, CURR));
  }
};
Tree w;
void input() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    inv[a[i]] = i;
  }
}
void solve() {
  w.init(1, 1, n);
  double ans = 0.0;
  double coef = 1.0 / (1.0 * n * (n + 1) / 2.0);
  for (int i = n; i >= 1; --i) {
    node ret = w.query(1, 1, n, 1, inv[i]);
    ans += ret.cnt;
    ans -= ret.val * (n - inv[i] + 1) / 2.0;
    w.update(1, 1, n, inv[i], coef * inv[i]);
  }
  w.init(1, 1, n);
  for (int i = 1; i <= n; ++i) {
    node ret = w.query(1, 1, n, 1, inv[i]);
    ans += ret.val * (n - inv[i] + 1) / 2.0;
    w.update(1, 1, n, inv[i], coef * inv[i]);
  }
  printf("%.12lf\n", ans);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}