Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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: Accepted)

class Array
  def merge_sort
    merge = -> l, r do
      ret = []
      for i in 0...size do
        case
        when l.empty?
          $cnt += l.size
          ret << r.shift
        when r.empty?
          ret << l.shift
        when l.first > r.first
          $cnt += l.size
          ret << r.shift
        else
          ret << l.shift
        end
      end
      ret
    end
    if size != 1
      mid = size / 2
      l = self[0...mid].merge_sort
      r = self[mid...size].merge_sort
      merge.(l, r)
    else
      self
    end
  end
end


$cnt = 0
gets
gets.split(' ').map(&:to_i).merge_sort
puts $cnt