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 Kotlin (judge verdict: Accepted)

fun main(args: Array<String>) {
    readLine()
    val list = readLine()!!.split(' ').map { it.toInt() }
    println(countInversion(list))
}

private fun countInversion(list: List<Int>): Long {

    var cnt = 0L

    val intArray = list.toIntArray()
    val leftArray = IntArray(list.size / 2 + 2)
    val rightArray = IntArray(list.size / 2 + 2)

    fun merge(left: Int, mid: Int ,right: Int) {
        val leftSize = mid - left
        for (i in 0 until leftSize) {
            leftArray[i] = intArray[left + i]
        }
        leftArray[leftSize] = Int.MAX_VALUE

        val rightSize = right - mid
        for (i in 0 until rightSize) {
            rightArray[i] = intArray[mid + i]
        }
        rightArray[rightSize] = Int.MAX_VALUE

        var l = 0
        var r = 0

        for (i in left until right) {
            if (leftArray[l] <= rightArray[r]) {
                intArray[i] = leftArray[l++]
            } else {
                cnt += leftSize - l
                intArray[i] = rightArray[r++]
            }
        }
    }

    fun sort(left: Int, right: Int) {
        if (left + 1 < right) {
            val mid = (left + right) / 2
            sort(left, mid)
            sort(mid, right)
            merge(left, mid, right)
        }
    }

    sort(0, list.size)

    return cnt
}