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: Accepted)

class Msort
  attr_reader :arr, :count, :tcount

  def initialize(arr)
    @arr = arr
    @count = 0
    @tcount = 0
  end

  def exec(array=arr)
    if array.length == 1
      array
    else
      mid = array.length / 2 - 1
      merge(exec(array[0..mid]), exec(array[mid+1..-1]))
    end
  end

  def merge(left, right)
    tl = left.shift
    tr = right.shift
    result = []
    loop do
      break if [tl, tr].all?(&:nil?)
      @count += 1
      if tl.nil?
        result << tr
        tr = right.shift
        next
      end
      if tr.nil?
        result << tl
        tl = left.shift
        next
      end
      if tl <= tr
        result << tl
        tl = left.shift
      else
        @tcount += left.length + 1
        result << tr
        tr = right.shift
      end
    end
    result
  end
end

n = gets.to_i
arr = gets.split.map(&:to_i)
m = Msort.new(arr)
m.exec
puts m.tcount