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: Wrong Answer)

def merge(a, left, mid, right)
  n1 = mid - left
  n2 = right - mid
  _L = a[left..mid-1]
  _R = a[mid..right-1]
  _L[n1] = _R[n2] = 1.0 / 0.0
  i = j = 0
  cnt = 0
  left.upto(right-1){|k|
    if _L[i] <= _R[j] then
      a[k] = _L[i]
      i += 1
    else
      a[k] = _R[j]
      j += 1
      cnt += 1
    end
  }
  return cnt
end

def mergeSort(a, left, right)
  count = 0
  if left+1 < right then
    mid = (left + right) / 2
    count += mergeSort(a, left, mid)
    count += mergeSort(a, mid, right)
    count += merge(a, left, mid, right)
  end
  return count
end

n = gets.to_i
a = gets.split.map(&:to_i)

count = mergeSort(a, 0, n)

#puts a.join(" ")
puts count - 1