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 Python (judge verdict: Time Limit Exceeded)

import array

def merge_sort(A):
    if len(A) <= 1:
        return A, 0
    mid = len(A) // 2
    left, lcount = merge_sort(A[:mid])
    right, rcount = merge_sort(A[mid:])
    merged, count = merge(left, right)
    return merged, lcount + rcount + count

def merge(left, right):
    lpos = 0
    rpos = 0
    rcount = 0
    merged = array.array('I', [])
    len_left = len(left)
    len_right = len(right)
    while lpos < len_left and rpos < len_right:
        if left[lpos] < right[rpos]:
            merged.append(left[lpos])
            lpos += 1
        else:
            merged.append(right[rpos])
            rpos += 1
            rcount += len(left[lpos:])
    
    if left[lpos:]:
        merged.extend(left[lpos:])
    if right[rpos:]:
        merged.extend(right[rpos:])
    
    return merged, rcount
    
if __name__ == '__main__':
    n = int(raw_input())
    A = array.array('I', map(int, raw_input().split()))
    
    sorted, count = merge_sort(A)
    #print ' '.join(str(x) for x in sorted)
    print count