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: Time Limit Exceeded)

import java.util.Scanner;

class Main {
	public static void main(String args[]) {
		Scanner scanner = new Scanner(System.in);
		int  num = scanner.nextInt();//要素数
		int array[] = new int[num];
		int count = 0;
		
		for(int i = 0; i < num; i ++ ){
			array[i] = scanner.nextInt();
		}
		
		count = bubbleSort(array, num);

		System.out.println(count);
		
		scanner.close();
	}

	public static int bubbleSort( int[] numArray, int num){ // N 個の要素を含む 0-オリジンの配列 A
		int count = 0;
		
		//バブルソート処理
		for (int i = 0; i < num - 1; i++) {
				for (int j = num - 1; j > i; j--) {
					if (numArray[j - 1] > numArray[j]) {
						int work = numArray[j - 1];
						numArray[j - 1] = numArray[j];
						numArray[j] = work;
						count++;
					}
				}
			}
		return count;
	}
}