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: Wrong Answer)

import java.io.IOException;
import java.util.Scanner;

class Main {
	static final int INF = 1000000001;

	static int changecount;
   public static void main(String args[]) throws IOException{

      Scanner sc = new Scanner(System.in);

	   int n = Integer.parseInt( sc.next()  );

	   int[] A = new int[n];

	   for(int i = 0;i < n; i++){

		   A[i] = Integer.parseInt(sc.next());
	   }

	    bubleSort(A, 0, n);

	   for(int i = 0;i < n; i++){
		   System.out.print(A[i]);
		   if(i < n-1 ){
			   System.out.print(" ");
		   }
	   }
	   System.out.println();
	   System.out.println(changecount);
	   sc.close();
   }



   public static void buble( 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];


	   for(int i = 0; i < n1;i++ ){
		   L[i] = A[left + i];

	   }

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

	   }

	   L[n1] =  INF;
	   R[n2] =  INF;

	   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;
			      changecount+=j;
		   }else{
			   A[k] = R[j];
			      j = j + 1;

		   }

	   }


   }


   public static void bubleSort(int[] A, int left, int right){


	   if(left+1 < right){
		    int mid = (left + right)/2;
		    bubleSort(A, left, mid);
		    bubleSort(A, mid, right);
		   buble(A, left, mid, right);

	   }



   }
}