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 = a[left..left+n1-1]
	r = a[mid..mid+n2-1]
	l[n1] = 10 ** 9
	r[n2] = 10 ** 9

	i = 0
	j = 0
	
	for k in left..right-1
		if l[i] < r[j]
			a[k] = l[i]
			i += 1
		else
			a[k] = r[j]
			j += 1
			
			$cnt += l.length - i - 1
		end
	end
end

def merge_sort(a, left, right)
	if left + 1 < right
		mid = (left + right) / 2
		merge_sort(a, left, mid)
		merge_sort(a, mid, right)
		merge(a, left, mid, right)
	end
end

n = gets.to_i
a = gets.split(" ").map(&:to_i)
$cnt = 0

merge_sort(a, 0, n)

puts $cnt