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 JAVA (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
	static long cnt = 0; // 反転数
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt(); // 要素の個数
		int[] num = new int[n]; // 個数のリスト

		for (int i = 0; i < n; i++) {
			num[i] = scan.nextInt();
		}
		mergeSort(num,0,n);
		System.out.println(cnt);
	}


	
	/**
	 * 比較してソートする
	 * @param A	ソートするリスト
	 * @param left	ソート対象の一番左側のリスト番号
	 * @param mid	分割地点
	 * @param right	ソート対象の一番右側のリスト番号
	 */
	public static void merge(int[] A,int left,int mid,int right){
		int n1 = mid -left;
		int n2 = right -mid;
		int[] L =new int[n1+1];
		int[] R =new int[n2+1];
		L[n1] = Integer.MAX_VALUE;//∞
		R[n2] = Integer.MAX_VALUE;//∞
		for(int i = 0;i<n1;i++){

			L[i] = A[left+i];
		}
		for(int i = 0;i<n2;i++){	  
			R[i] = A[mid+i];
		}

		int i = 0;
		int j = 0;
		for(int k = left; k<right;k++ ){
			
			if(L[i] <= R[j]){
				A[k] = L[i];
				i = i+1;
			}else{
				if (i < n1) {
                    cnt += n1 - i;
                }

				A[k] = R[j];
				j = j+1;
			}

		}


	}
	/**
	 *  マージソート
	 * @param A	ソートするリスト
	 * @param left　ソート対象の一番左側のリスト番号
	 * @param right　ソート対象の一番右側のリスト番号
	 * @return
	 */
	private static  void  mergeSort(int[] A, int left, int right){
		if(left +1 < right){

			int mid = (left + right) /2;
			mergeSort(A,left,mid);
			mergeSort(A,mid , right);
			merge(A,left,mid,right);
		}
	}
}