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 Python3 (judge verdict: Accepted)

import sys
cnt = 0

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

def merge(a, left, mid, right):
	n1 = mid - left
	n2 = right - mid
	L = a[left:left+n1]
	R = a[mid:mid+n2]
	L.append(sys.maxsize)
	R.append(sys.maxsize)
	i = j = 0
	tmp = n1
	for k in range(left, right):
		if L[i] <= R[j]:
			if R[j] != sys.maxsize:
				tmp -= 1
			a[k] = L[i]
			i += 1
		else:
			if L[i] != sys.maxsize:
				global cnt
				cnt += tmp
			a[k] = R[j]
			j += 1		
	
n = int(input())
a = list(map(int, input().split()))
merge_sort(a, 0, len(a))
print(cnt)