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 <cstdio>

using namespace std;

#define MAX 200000
#define Sentinel 2000000000
typedef long long llong;

int L[ MAX / 2 + 2], R[ MAX / 2 + 2];

llong cnt = 0;

void merge( int A[], int left, int mid, int right){

    int i , j , k;

    int n1 = mid - left;
    int 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] = R[n2] = Sentinel;

    i = 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 += 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 A[MAX], n, i;

    scanf("%d", &n);

    for( i = 0 ; i < n ; i++ ) scanf( "%d", &A[i] );


    mergesort( A, 0, n);

    printf("%lld\n", cnt);

    return 0;

}