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(a, left, mid, right)
  n1 = mid - left
  n2 = right - mid
  l = (0..n1).to_a
  r = (0..n2).to_a
  0.upto(n1-1).each {|i| l[i] = a[left+i]}
  0.upto(n2-1).each {|i| r[i] = a[mid+i]}
  l[n1] = Float::INFINITY
  r[n2] = Float::INFINITY
  i = 0
  j = 0
  count = 0
  left.upto(right - 1).each {|k|
    if l[i] <= r[j]
      a[k] = l[i]
      i += 1
    else
      a[k] = r[j]
      j += 1
      count += n1 - i
    end
  }
  count
end

def merge_sort(a, left, right)
  return 0 unless left + 1 < right
  mid = (left + right) / 2
  count = merge_sort(a, left, mid)
  count += merge_sort(a, mid, right)
  count += merge(a, left, mid, right)
  count
end

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

count = merge_sort(s, 0, n)
puts count