Problem:
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The Number of Inversions

For a given sequence $A = \{a_0, a_1, ... a_{n-1}\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:

bubbleSort(A)
cnt = 0 // the number of inversions
for i = 0 to A.length-1
for j = A.length-1 downto i+1
if A[j] < A[j-1]
swap(A[j], A[j-1])
cnt++

return cnt

For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.

Input

In the first line, an integer $n$, the number of elements in $A$, is given. In the second line, the elements $a_i$ ($i = 0, 1, .. n-1$) are given separated by space characters.

output

Print the number of inversions in a line.

Constraints

$ 1 \leq n \leq 200,000$

$ 0 \leq a_i \leq 10^9$

$a_i$ are all different

Sample Input 1

5
3 5 2 1 4

Sample Output 1

6

Sample Input 2

3
3 1 2

Sample Output 2

2

Solution in OCaml (judge verdict: Accepted)

let cnt = ref 0

let merge a l m r =
  let ni = m - l in
  let init_ary n p =
    Array.init (n + 1) (fun i -> if i = n then max_int else a.(p + i))
  in
  let ai = init_ary ni l in
  let aj = init_ary (r - m) m in
  let rec doit i j k =
    if k < r then begin
      if ai.(i) <= aj.(j) then (a.(k) <- ai.(i); doit (i + 1) j (k + 1))
      else (a.(k) <- aj.(j); cnt := !cnt + (ni - i); doit i (j + 1) (k + 1))
    end
  in
  doit 0 0 l

let rec merge_sort a l r =
  if (l + 1) < r then begin
    let m = (l + r) / 2 in
    merge_sort a l m;
    merge_sort a m r;
    merge a l m r
  end

let () =
  let n = read_int () in
  let a = Array.init n (fun _ -> Scanf.scanf "%d " (fun i -> i)) in
  merge_sort a 0 n;
  Printf.printf "%d\n" !cnt