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 MAX 500000
#define BUMP 10000000000

long long count = 0;
int L[MAX / 2 + 2], R[MAX / 2 + 2];

void marge(int A[], int left, int mid, int right){
    int n1 = mid - left;
    int n2 = right - mid;
    int i, j, m;
   
    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] = BUMP;
    for(i = 0, j = 0, m = left; m < right; m++){
        if(L[i] <= R[j]){
            A[m] =L[i];
            i++;
        }else{
            A[m] = R[j];
            j++;
            count += n1 - i ;
        }
    }
}

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

int main(void){
    int A[MAX];
    int n, i;

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

    margesort(A, 0, n);

    printf("%ld\n",count);
    return 0;
}