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 Ruby (judge verdict: Wrong Answer)

$count = 0
def merge(left_array, right_array)
  return_array = []
  left_index, right_index = 0, 0
  left_array << -1
  right_array << -1
  0.upto(left_array.length + right_array.length - 3) do
    $count += 1
    if left_array[left_index] > right_array[right_index]
      return_array << left_array[left_index]
      left_index += 1
    else
      return_array << right_array[right_index]
      right_index += 1
    end
  end
  return_array
end

def merge_sort(target_array)
  right = target_array.length
  mid = target_array.length / 2
  left_array = target_array[0...mid]
  right_array = target_array[mid...right]
  return target_array if target_array.length == 1
  return merge(merge_sort(left_array), merge_sort(right_array))
end

gets.to_i
input = gets.split(' ').map(&:to_i)
puts merge_sort(input).join(' '), $count