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: Runtime Error)

object Main extends App {
  import scala.collection.mutable.ArrayBuffer

  val n = io.StdIn.readLine().toInt
  val arr: Array[Int] = split(io.StdIn.readLine())

  def split(str: String): Array[Int] = {
    val buf: ArrayBuffer[Int] = ArrayBuffer.empty[Int]
    var num_builder = 0
    val iter = str.toIterator

    while(iter.hasNext) {
      val v = iter.next()
      if (v != ' ') num_builder = num_builder*10 + v.toInt
      else {buf += num_builder; num_builder = 0}
    }
    (buf += num_builder).toArray
  }

  def sum(index: Int, acc: Int = 0): Int =
    if (index > 0) sum(index - (index & (-index)), acc + bit(index))
    else acc

  def update(n: Int, index: Int, v: Int): Unit =
    if (index <= n) {
      bit(index) += v
      update(n, index + (index & -index), v)
    }

  var count: Long = 0
  val maxVal = arr.max
  val bit: Array[Int] = Array.fill(maxVal+1)(0)
  for(i <- n-1 to 0 by -1) {
    count += sum(arr(i)-1)
    update(maxVal, arr(i), 1)
  }

  println(count)
}