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)

def partition(A, p, r):
    x = A[r][1]
    i = p-1
    for j in range(p, r):
        if A[j][1] <= x:
            i = i + 1
            (A[i], A[j]) = (A[j], A[i])

    (A[i+1], A[r]) = (A[r], A[i+1])

    return i+1
    
def quicksort(A, p, r):
    if p < r:
        q = partition(A, p, r)
        quicksort(A, p, q-1)
        quicksort(A, q+1, r)

def stable(A, B):
    for i in range(len(A)):
        if A[i] != B[i]:
            return "Not stable"
    return "Stable"

def merge(A, left, mid, right):
    count = 0
    n1 = mid - left
    n2 = right - mid
    L = []
    R = []

    for i in range(n1):
        L.append(A[left+i])
    for i in range(n2):
        R.append(A[mid+i])

    L.append(10**9)
    R.append(10**9)

    i = 0
    j = 0

    for k in range(left, right):
        if L[i] <= R[j]:
            A[k] = L[i]
            i = i + 1
        else:
            A[k] = R[j]
            j = j + 1
            count = count + n1 - i
    return count

def mergeSort(A, left, right):
    if left + 1 < right:
        mid = (left + right)//2
        return mergeSort(A, left, mid) + mergeSort(A, mid, right) + merge(A, left, mid, right)
    else:
        return 0

n = int(input())
X = list(map(int, input().split()))
x = mergeSort(X, 0, n)

print(x)