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 C (judge verdict: Accepted)

#include <stdio.h>

#define M 100001
#define N 200000
#define INFTY 1000000000
long long count;

void merge(int A[],int left,int  mid, int  right)
{
	int i,j,k,n1,n2,L[M],R[M];
 	n1 = mid - left;
  	n2 = right - mid;
  
	
	for (i = 0; i<n1; i++) 
	{
		L[i]=A[left+i];
	}
	for(i = 0; i<n2; i++)
	{
		R[i]=A[mid+i];
 
	}

  	L[n1] = INFTY;
  	R[n2] = INFTY;
  	i = 0;
  	j = 0;
  	
	for (k = left; k<right; k++)
	{
		if (L[i] <= R[j]) 
		{
			A[k] = L[i];
      		i++;
		}
    else 
		{
			A[k] = R[j];
      		j++;
			count += n1-i;
		}
	}
}


void mergeSort(int A[],int left,int right)
{
	int mid;
	if (left+1 < right)
	{
    	mid = (left + right)/2;
    	mergeSort(A,left, mid);
    	mergeSort(A,mid, right);
    	merge(A,left, mid, right);
	}
}


int main()
{
	
	int n,i,A[N];
	count = 0;
	scanf("%d",&n);
	for(i = 0; i< n; i++)
	{
		scanf("%d",&A[i]);
	}
	
	mergeSort(A,0,n);
	
	 printf("%ld\n", count);   
	return 0;
}