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)

gets
arr = gets.split.map(&:to_i)

@count = 0

def merge_sort(arr)
  return arr if arr.size <= 1

  mid = arr.size / 2
  left = arr[0..(mid - 1)]
  right = arr[mid..(arr.size - 1)]

  merge(merge_sort(left), merge_sort(right))
end

def merge(left, right)
  result = []
  until left.empty? || right.empty?
    if left[0] < right[0]
      result << left.shift
    else
      # 右の要素が左の残りの要素より小さい場合、
      # 左の要素の残りの数だけ反転して、ソート済みの配列に並ぶ
      @count += left.size
      result << right.shift
    end
  end

  result << left.shift until left.empty?
  result << right.shift until right.empty?
  result
end

merge_sort(arr)
puts @count