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 Solver
  @@count = 0
  @@a

  def self.init_a(a)
    @@a = a
  end

  def self.merge(left, mid, right)
    n1 = mid - left
    n2 = right - mid
    l = @@a[left..mid]
    r = []

    (0..n2-1).each do |i|
      r[i] = @@a[mid + i]
    end
  
    l[n1] = Float::INFINITY
    r[n2] = Float::INFINITY
    i = 0
    j = 0
    (left..right-1).each do |k|
      if l[i] <= r[j]
        @@a[k] = l[i]
        i = i + 1
      else
        @@count += l.size - (i+1)
        @@a[k] = r[j]
        j = j + 1
      end
    end
  end
  
  def self.mergeSort(left, right)
    if left+1 < right
      mid = (left + right)/2
      mergeSort(left, mid)
      mergeSort(mid, right)
      merge(left, mid, right)
    end
  end

  def self.puts_result
    puts @@count
  end
end

n = gets.to_i
a = gets.split(' ').map(&:to_i)
Solver.init_a(a)
Solver.mergeSort(0, n)
Solver.puts_result