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 MIN_N 1
#define MAX_N 200000
#define MIN_A 0
#define MAX_A 1000000000
  
int array[MAX_N];
long invcnt = 0;
  
void merge(int *array, int left, int mid, int right)
{
    int n1 = mid - left;
    int n2 = right - mid;
    int i, j, k, *left_array, *right_array;
  
    left_array = malloc(sizeof(int) * n1 + 100);
    right_array = malloc(sizeof(int) * n2 + 100);
  
    for(i = 0; i < n1; i++) 
        left_array[i] = array[left + i];
    for(i = 0; i < n2; i++)
        right_array[i] = array[mid + i];
    left_array[n1] = MAX_A + 1;
    right_array[n2] = MAX_A + 1;
    i = 0;
    j = 0;
    for(k = left; k < right; k++) {
        if(left_array[i] <= right_array[j]) 
            array[k] = left_array[i++];
        else {
            invcnt += n1 - i;
            array[k] = right_array[j++];
        }
          
    }
    free(left_array);
    free(right_array);
}
  
void merge_sort(int *array, int left, int right)
{
    if(left + 1 < right) {
        int mid;
  
        mid = (left + right) / 2;
        merge_sort(array, left, mid);
        merge_sort(array, mid, right);
        merge(array, left, mid, right);
    }
  
}
  
int main(void)
{
    int i, n;
  
    scanf("%d", &n);
    for(i = 0; i < n; i++)
    {
        int a;
        scanf("%d", &a);
        array[i] = a;
    }
    merge_sort(array, 0, n);
    printf("%ld\n", invcnt);
  
    return 0;
}