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: Time Limit Exceeded)

class Array
  def swap!(a, b)
    self[a], self[b] = self[b], self[a]
    self
  end
end

def reverse_bubble_sort(array)
  isSorted = false
  count = 0
  index = 0
  until isSorted do
    swaped = false
    (array.length - 1).downto(index) do |comp_index|
      break if comp_index.zero?
      if array[comp_index] < array[comp_index - 1]
        array.swap! comp_index, (comp_index - 1)
        count += 1
        swaped = true
      end
    end
    isSorted = true unless swaped
    index += 1
  end
  count
end
n = gets.to_i
array = gets.split(' ').map(&:to_i)
count = reverse_bubble_sort array
puts count