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 INFTY 1000000000

long long cnt; //intでは扱えないためlong long(巨大な数)
long long merge(int A[],int left,int mid,int right){
	int n1=mid-left;		//左側の要素数
	int n2=right-mid;		//右側の要素数
	cnt=0;
	int L[n1],R[n2];
	int i,j,k;		
	for(i=0;i<n1;i++){		//左半分の要素代入
		L[i]=A[left+i];
	}
	for(i=0;i<n2;i++){		//右半分の要素代入
		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++;
			cnt=cnt+n1-i;
		}
	}
	return cnt;
}

long long mergeSort(int A[],int left,int right){
	int mid;
	long long a,b,c;	//intでは扱えないためlong long(巨大な数)
	if(right-left>1){	//両端の交差の有無で判断
		mid=(left+right)/2;
		a=mergeSort(A,left,mid);
		b=mergeSort(A,mid,right);
		c=merge(A,left,mid,right);
		return a+b+c;
	}
	return 0;
}

int main(void){
	
	int n;
	scanf("%d",&n);
	int A[n];
	int i;
	long long ans;

	
	for(i=0;i<n;i++){
		scanf("%d",&A[i]);
	}
	
	ans=mergeSort(A,0,n);
	printf("%lld\n",ans);
	
	return 0;
	
}