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: Accepted)

import scala.io.StdIn.{readLine, readInt}

object Main extends App {
  import MergeSort.mergeSort

  readInt()
  val xs  = readLine().split(' ').map { _.toLong }
  val ans = mergeSort(xs)
  println(ans)
}

object MergeSort {
  val Max: Int      = 200000
  val Sentinel: Int = 2000000000

  def mergeSort(xs: Array[Long]): Long = {

    val ls = new Array[Long](Max / 2 + 2)
    val rs = new Array[Long](Max / 2 + 2)

    def run(left: Int, right: Int): Long = {
      val m = right - left
      if (m <= 1) 0
      else {
        val mid  = (left + right) / 2
        val cnt1 = run(left, mid)
        val cnt2 = run(mid, right)
        val cnt3 = merge(left, mid, right)
        cnt1 + cnt2 + cnt3
      }
    }

    def merge(left: Int, mid: Int, right: Int): Long = {
      val n1   = mid - left
      val n2   = right - mid
      Array.copy(xs, left, ls, 0, n1)
      Array.copy(xs, mid, rs, 0, n2)
      ls(n1) = Sentinel
      rs(n2) = Sentinel
      var i, j = 0
      var cnt  = 0L
      var k    = left
      while (k < right) {
        if (ls(i) <= rs(j)) {
          xs(k) = ls(i)
          i += 1
        } else {
          xs(k) = rs(j)
          j += 1
          cnt += n1 - i
        }
        k += 1
      }

      cnt
    }

    run(0, xs.length)
  }
}