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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
int msort(int, int, int*);
main()
{
        int i, n, *a;
        scanf("%d", &n);
        a = (int*)malloc(sizeof(int) * n);
        for (i = 0; i < n; i++) {
                scanf("%d", &a[i]);
        }
        printf("%d\n", msort(0, n - 1, a));
        free(a);
}
int msort(int left, int right, int *a)
{
        int i, j, center, num, base, reverse = 0, work[right - left + 1];
        if (left >= right) return 0;
        center = (left + right) / 2;
        reverse += msort(left, center, a);
        reverse += msort(center + 1, right, a);
        for (num = 0, i = left; i <= center; num++, i++) {
                work[num] = a[i];
        }
        j = 0, base = left;
        while (i <= right && j < num) {
                if (a[i] < work[j]) {
                        a[base++] = a[i++];
                        reverse += i - base;
                } else {
                        a[base++] = work[j++];
                }
        }
        while (j < num) {
                a[base++] = work[j++];
        }
        return reverse;
}