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: Time Limit Exceeded)

#include<stdio.h>
#define MAX 200000

int main() {
	int N;
	int A[MAX];
	long long cnt = 0;
	int sp, buff;
	scanf("%d", &N);
	for (int i = 0; i < N; i++) {
		scanf("%d", &A[i]);
	}
	for (int i = 0; i < N; i++) {
		sp = i;
		buff = A[i];
		while (sp != 0 && A[sp - 1] > buff) {
			A[sp] = A[sp - 1];
			sp--;
			cnt++;
		}
		A[sp] = buff;
	}

	printf("%lld\n", cnt);

	return 0;
}