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 Scala (judge verdict: Wrong Answer)

object Main {
    var res = 0
    def msort(arr: Array[Int]): Array[Int] = {
        if (arr.size > 1) {
            val a = msort(arr.take(arr.size/2))
            val b = msort(arr.drop(arr.size/2))
            var i, j, k = 0
            var f = true
            while (f) {
                if (i == a.size && j == b.size) f = false
                else if (j == b.size) {arr(k) = a(i); i += 1; k += 1}
                else if (i == a.size) {arr(k) = b(j); j += 1; k += 1}
                else if (a(i) < b(j)) {arr(k) = a(i); i += 1; k += 1}
                else {arr(k) = b(j); j += 1; k += 1; res += (a.size-i)}
            }
        }
        arr
    }
    def main(args: Array[String]) = {
        val in = io.Source.stdin.getLines.map(_.split(" ").map(_.toInt))
        in.next
        msort(in.next)
        println(res)
    }
}