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>
#include <stdlib.h>
#define iMAX 1000000000

long long marge ( long *A, long left, long mid, long right)
{
    long d1,d2,i,j,k;
    long long cnt=0l;
    d1 = mid - left;
    d2 = right - mid;
    long *L = malloc(sizeof(long long)*(d1+1));
    long *R = malloc(sizeof(long long)*(d2+1));
    
    for( i=0l; i<d1; i++ ) L[i] = A[left+i];
    for( i=0l; i<d2; i++ ) R[i] = A[mid+i];

    L[d1] = iMAX;
    R[d2] = iMAX;

    for( k=left,i=0l,j=0l; k<right; k++ )
    {
        if( L[i] <= R[j] )
        {
            A[k] = L[i++];
        }else{
            A[k] = R[j++];
            cnt += d1-i;
        }
    }
    
    free(L);
    free(R);
    return cnt;
}


long long margeSort( long  *A, long  left,  long right)
{
    long mid;
    long long cnt=0l;
    
    if( left+1l < right )
    {
        mid = ( left + right ) /2l;
        cnt += margeSort( A, left, mid );
        cnt += margeSort( A, mid, right );
        cnt += marge( A, left, mid, right );
    }
    
    return cnt;
}


int main()
{
    long n,i;
    scanf( "%ld", &n );
    long *S = malloc(sizeof(long)*n);
    
    for( i=0l; i<n; i++ ){
        scanf( "%ld", &S[i] );
    }
    
    printf( "%lld\n",margeSort( S, 0l, n ));
    free(S);
    return 0;
}