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)

object Main extends App {
  var count: Long = 0
  val n = io.StdIn.readLine()
  val arr: collection.mutable.ArrayOps[Int] = io.StdIn.readLine().split(' ') map (_.toInt)

  def merge(left: Int, middle: Int, right: Int): Unit = {
    val l = arr.slice(left, middle)
    val r = arr.slice(middle, right)
    val lLen = l.length
    val rLen = r.length

    var i = left
    var li = 0
    var ri = 0

    while(li < lLen && ri < rLen) {
      if (l(li) <= r(ri)) {
        arr(i) = l(li)
        i += 1
        li += 1
      } else {
        arr(i) = r(ri)
        count += (lLen - li)
        i += 1
        ri += 1
      }
    }

    if (li < lLen) while (li < lLen) {arr(i) = l(li); i += 1; li += 1}
    else while (ri < rLen) {arr(i) = r(ri); i += 1; ri += 1}
  }

  def msort(left: Int, right: Int): Unit = {
    val middle = (left+right+1)/2
    if (middle < right) {
      msort(left, middle)
      msort(middle, right)
      merge(left, middle, right)
    } else {
      ()
    }
  }

  msort(0, arr.length)
  println(count)
}