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)

n = int(input())
lis = list(map(int,input().split()))
def merge_sort(arr):
    if len(arr) <= 1:
        return arr

    mid = len(arr) // 2
    # ここで分割を行う
    left = arr[:mid]
    right = arr[mid:]

    # 再帰的に分割を行う
    left = merge_sort(left)
    right = merge_sort(right)

    # returnが返ってきたら、結合を行い、結合したものを次に渡す
    return merge(left, right)


def merge(left, right):
    global ans
    merged = []
    l_i, r_i = 0, 0

    # ソート済み配列をマージするため、それぞれ左から見ていくだけで良い
    while l_i < len(left) and r_i < len(right):
        # ここで=をつけることで安定性を保っている
        if left[l_i] <= right[r_i]:
            merged.append(left[l_i])
            l_i += 1
        else:
            merged.append(right[r_i])
            r_i += 1
            ans += (len(left) - l_i)

    # 上のwhile文のどちらかがFalseになった場合終了するため、あまりをextendする
    if l_i < len(left):
        merged.extend(left[l_i:])
    if r_i < len(right):
        merged.extend(right[r_i:])
    return merged

ans = 0
merge_sort(lis)
print(ans)