Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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: Time Limit Exceeded)

def bubble_sort_revision(target_list):

    list_length = len(target_list)
    
    flag = True
    change_count = 0
    
    for top_index in range(1,list_length):
        for i in range(top_index, list_length)[::-1]:
            if target_list[i] < target_list[i - 1] and target_list[top_index] >= target_list[i]:
                tmp = target_list[i]
                target_list[i] = target_list[i - 1]
                target_list[i - 1] = tmp
                change_count += 1
                
    return change_count
def solve(n, target_list):
    a = [0 for i in range(max(target_list) + 1)]
    ans = 0
    for i in range(n):
        ans += i - sum(a[:target_list[i]])
        a[target_list[i]] += 1
    return ans

if __name__ == "__main__":
    l = int(input())
    target_list = [int(i) for i in input().split()]
    print(solve(l, target_list))