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
  @@cnt = 0

  def cnt
    c = @@cnt
    @@cnt = 0
    c
  end

  def merge_sort
    if size != 1
      mid = size / 2
      l = self[0...mid].merge_sort
      r = self[mid...size].merge_sort
      merge(l, r)
    end
    self
  end

  private def merge l, r
    i = 0
    b = dup
    size.times do
      case
      when l.empty?
        @@cnt += b.index(r.first) - i
        self[i] = r.shift
      when r.empty?
        self[i] = l.shift
      when r.first < l.first
        @@cnt += b.index(r.first) - i
        self[i] = r.shift
      else
        self[i] = l.shift
      end
      i += 1
    end
    self
  end
end

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