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)

from _collections import deque

SENIAL = 1000000000
def mergeSort(A,left,right):
    if left + 1 < right:
        middle = (left + right) // 2
        mergeSort(A, left, middle)
        mergeSort(A, middle, right)
        merge(A,left,right,middle)
    
def merge(A,left,right,middle):
    global count        
    L = deque(A[left:middle])
    R = deque(A[middle:right])
    L.append(SENIAL)
    R.append(SENIAL)
    for i in range(left,right):
        if L[0] < R[0]:
            A[i] = L.popleft()
        else:
            A[i] = R.popleft()
            count += len(L) -1 
count = 0     
input()
A = list(map(int,input().split()))
mergeSort(A,0,len(A))
print(count)