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 math

def merge(A, left, mid, right, cnt):
    n1 = mid - left
    n2 = right - mid
    L = [0]*(n1+1)
    R = [0]*(n2+1)
    for i in range(n1):
        L[i] = A[left + i]
    for i in range(n2):
        R[i] = A[mid + i]
    L[n1] = math.inf
    R[n2] = math.inf
    i = 0
    j = 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
            cnt += mid - left - i
            
    return A, cnt


def mergeSort(A, left, right, cnt):
    if left+1 < right:
        mid = (left + right)//2
        A, cnt = mergeSort(A, left, mid, cnt)
        A, cnt = mergeSort(A, mid, right, cnt)
        A, cnt = merge(A, left, mid, right, cnt)
    return A, cnt
        
n = int(input())
A= list(map(int,input().split()))

A_out, cnt_out = mergeSort(A,0,len(A),0)

print(str(cnt_out))