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>
#include <limits.h>

void mergeSort(int* list, int left, int right);

unsigned long count = 0;

int main() {
	int n, * a;
	scanf("%d", &n);
	a = (int*)malloc(n * sizeof(int));
	for (int i = 0; i < n; i++)
		scanf("%d", a + i);
	
	mergeSort(a, 0, n);

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

void mergeSort(int* list, int left, int right) {
	if (left + 1 < right) {
		int middle = (left + right) / 2;
		mergeSort(list, left, middle);
		mergeSort(list, middle, right);

		int nL = middle - left,
			nR = right - middle,
			*l = malloc((nL + 1) * sizeof(int)),
			*r = malloc((nR + 1) * sizeof(int));
		for (int i = 0; i < nL; i++)
			l[i] = list[left + i];
		for (int i = 0; i < nR; i++)
			r[i] = list[middle + i];
		l[nL] = INT_MAX, r[nR] = INT_MAX;
		int i = 0, j = 0;
		for (int k = left; k < right; k++)
			if (l[i] > r[j])
				list[k] = r[j++], count += nL - i;
			else
				list[k] = l[i++];
		free(l);
		free(r);
	}
}