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 Go (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"io"
	"os"
	"strconv"
)

const (
	SENTINEL = 2000000000
)

func main() {
	fmt.Println(answer(os.Stdin))
}

func answer(r io.Reader) int {
	sc := bufio.NewScanner(r)
	sc.Split(bufio.ScanWords)
	sc.Scan()
	n, _ := strconv.Atoi(sc.Text())
	A := make([]int, n)
	for i := 0; i < n; i++ {
		sc.Scan()
		A[i], _ = strconv.Atoi(sc.Text())
	}
	return mergeSort(A, n, 0, n)
}

func merge(A []int, n, left, mid, right int) int {
	n1 := mid - left
	n2 := right - mid
	L, R := make([]int, n1+1), make([]int, n2+1)
	for i := 0; i < n1; i++ {
		L[i] = A[left+i]
	}
	for i := 0; i < n2; i++ {
		R[i] = A[mid+i]
	}
	L[n1], R[n2] = SENTINEL, SENTINEL

	cnt := 0
	i, j := 0, 0
	for k := left; k < right; k++ {
		if L[i] <= R[j] {
			A[k] = L[i]
			i++
			continue
		}
		A[k] = R[j]
		j++
		cnt += n1 - i
	}
	return cnt
}

func mergeSort(A []int, n, left, right int) int {
	if left+1 < right {
		mid := (left + right) / 2
		v1 := mergeSort(A, n, left, mid)
		v2 := mergeSort(A, n, mid, right)
		v3 := merge(A, n, left, mid, right)
		return v1 + v2 + v3
	}
	return 0
}