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)

class BinaryIndexedTree(object):
    __slots__ = ["tree", "size"]

    def __init__(self, size: int):
        self.tree = [0]*(size+1)
        self.size = size+1

    def add(self, index: int, value: int):
        tree = self.tree

        while index < len(tree):
            tree[index] += value
            index += index & -index

    def sum(self, index: int):
        tree, result = self.tree, 0

        while index:
            result += tree[index]
            index -= index & -index

        return result


def solve(A):
    bit = BinaryIndexedTree(len(A)+1)
    ans = 0
    for i, n in enumerate(A):
        ans += i - bit.sum(n-1)
        bit.add(n, 1)

    return ans


if __name__ == "__main__":
    input()
    a = list(map(int, input().split()))
    d = {n: i for i, n in enumerate(sorted(a), start=1)}
    a = [d[n] for n in a]
    print(solve(a))