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>
#include <memory.h>

int bit[200005];

int sum(int i) {
	int s = 0;
	while(i > 0) {
		s += bit[i];
		i -= (i & -i);
	}
	return s;
}

void add(int i, int x, int n) {
	while(i<=n) {
		bit[i] += x;
		i += (i & -i);
	}
}

int main(void) {
	int i, n, a, ans = 0;
	scanf("%d\n", &n);
	
	memset(bit, 0x00, sizeof(bit));
	for(i=0;i<n;i++) {
		scanf("%d", &a);
		ans += (i - sum(a));
		add(a, 1, n);
	}
	
	printf("%d\n", ans);

	return 0;
}