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 BinaryIndexTree

  def initialize n, arr = [], init = 0, &block
    block = ->(a, b) { a + b } unless block_given?
    @fun = block

    @n = n
    @init = init

    @tree = Array.new(n+1) { init }
    arr.each.with_index(1) do |e, i|
      put(i, e)
    end
  end

  def put i, e
    while i <= @n
      @tree[i] = @fun.(@tree[i], e)
      i += i & -i
    end
  end

  def query i
    r = @init
    while i > 0
      r = @fun.(r, @tree[i])
      i -= i & -i
    end
    r
  end
end

n = gets.to_i
as = gets.split.map(&:to_i)
cs = as.sort.each.with_index(1).reduce({}) do |h, (a, i)|
  h[a] = i
  h
end

r = 0
bit = BinaryIndexTree.new(n)
as.each.with_index do |a, i|
  j = cs[a]
  r += i - bit.query(j)
  bit.put(j, 1)
end
puts r