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)

def merge_sort(arr, left, right)
  return 0 unless left + 1 < right

  count = 0

  mid = (left + right) / 2

  count += merge_sort(arr, left, mid)
  count += merge_sort(arr, mid, right)
  count += merge(arr, left, mid, right)

  count
end

def merge(arr, left, mid, right)
  count = 0

  n1 = mid - left
  n2 = right - mid

  left_arr = arr[left, n1]
  right_arr = arr[mid, n2]

  left_arr << Float::INFINITY
  right_arr << Float::INFINITY

  i = j = 0
  k = left

  while k < right
    if left_arr[i] <= right_arr[j]
      arr[k] = left_arr[i]
      i += 1
    else
      arr[k] = right_arr[j]
      j += 1
      count += n1 - i
    end

    k += 1
  end

  count
end

if $PROGRAM_NAME == __FILE__
  n = gets.to_i

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

  puts merge_sort(arr, 0, n)
end