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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
class Main {
   public static void main(String[] args)throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		int[] A = new int[n];
		
		long cnt;
		
		String[] str = br.readLine().split(" ");
		
		for(int i = 0; i < n; i++){
			A[i] = Integer.parseInt(str[i]);
		}
		
		cnt = Recursion(A);
		
		System.out.println(cnt);
	}
	
	static long Recursion(int data[]){
		long c = 0;
		
		for(int i = 0; i < data.length - 1; i++){
			for(int j = i + 1; j < data.length; j++){
				if(data[i] > data[j]){
					c++;
				}
			}
		}
		
		return c;
	}
}