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)

# Recursion / Divide and Conquer - The Number of Inversions
def conq(A,left,mid,right):
    global c
    n1 = mid - left
    n2 = right - mid
    L = A[left:mid] + [10**9+1]
    R = A[mid:right] + [10**9+1]
    i,j = 0,0
    for k in range(left,right):
        if L[i] <= R[j]:
            A[k] = L[i]
            i += 1
        else:
            A[k] = R[j]
            j += 1
            c += n1 - i

def num_inv(A,left,right):
    if left + 1 < right:
        mid = (left + right) // 2
        num_inv(A,left,mid)
        num_inv(A,mid,right)
        conq(A,left,mid,right)

n = int(input())
A = list(map(int, input().split()))
c = 0
num_inv(A,0,n)
print(c)