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: Memory Limit Exceeded)

object Main extends App {
  var count = 0

  def merge(a: Vector[Int], b: Vector[Int]): Vector[Int] = (a,b) match {
    case (_, Vector()) => a
    case (Vector(), _) => b
    case (ah+:at, bh+:bt) =>
      if (ah <= bh) ah +: merge(at, b)
      else {count += a.length; bh +: merge(a, bt)}
  }

  def msort(l: Vector[Int]): Vector[Int] = {
    if (l.length > 1) {
      val (x,y) = l.splitAt(l.length/2)
      val xx = msort(x)
      val yy = msort(y)
      merge(xx,yy)
    } else {
      l
    }
  }

  val n = io.StdIn.readLine()
  val l = io.StdIn.readLine().split(' ').toVector map (_.toInt)
  msort(l)
  println(count)
}