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)

INF = 2000000000
def merge(a, left, mid, right)
  cnt = 0
  n1 = mid - left
  n2 = right - mid
  l = []
  r = []
  (n1).times do |i|
    l[i] = a[left + i]
  end
  (n2).times do |i|
    r[i] = a[mid + i]
  end
  l[n1] = r[n2] = INF
  i = 0
  j = 0
  for k in left...right do
    if l[i] <= r[j]
      a[k] = l[i]
      i = i + 1
    else
      a[k] = r[j]
      j = j + 1
      cnt += n1 - i
    end
  end
  return cnt
end

def mergeSort(a, left, right)
  if left + 1 < right
    mid = (left + right) / 2
    v1 = mergeSort(a, left, mid)
    v2 = mergeSort(a, mid, right)
    v3 = merge(a, left, mid, right)
    return v1 + v2 + v3
  else
    return 0
  end
end

n = gets.to_i
a = gets.chomp.split(' ').map(&:to_i)
puts mergeSort(a, 0, a.size())