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: Wrong Answer)

# 16D8101012J ItoJun dj Python3
count=0
def merge(A, left, mid, right):
    global count
    h=[]
    L=A[left:mid]+[2**32]
    R=A[mid:right]+[2**32]
    i=0
    j=0
    for k in range(left,right):
        if L[i] <= R[j]:
            print("{}<={}".format(L[i],R[j]))
            A[k] = L[i]
            h.append(L[i])
            i += 1
        else:
            print("{}>{}".format(L[i],R[j]))
            A[k] = R[j]
            h.append(R[j])
            j += 1
            if i < len(L)-1:
                count += len(L) - 1 - i
        
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)

if __name__ == "__main__":
    n=int(input())
    A=list(map(int,input().split()))
    merge_sort(A,0,n)
    print(count)