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 <stdlib.h>

typedef struct _Node {
	int data;
	int count;
	struct _Node *left;
	struct _Node *right;
} Node;

static void setNode(Node *p, const int *data, Node *left, Node *right) {
	p->data = *data;
	p->count = 0;
	p->left = left;
	p->right = right;
}

Node *add(Node *p, const int *data, int *num) {
	int cmp;
	if (p == NULL) {
		p = (Node *)malloc(sizeof(Node));
		setNode(p,data,NULL,NULL);
	} else if ((cmp = *data - p->data) < 0) {
		*num += p->count + 1;
		p->left = add(p->left, data, num);
	}	else if (cmp > 0) {
		p->count++;
		p->right = add(p->right, data, num);
	}
	return p;
}

void terminate(Node *p) {
	if (p != NULL) {
		terminate(p->left);
		terminate(p->right);
		free(p);
	}
}

int main() {
	int n, i, data, numOfinv;
	Node *root = NULL;

	numOfinv = 0;
	scanf("%d",&n);
	for (i = 0; i < n; i++) {
		scanf("%d",&data);
		root = add(root,&data,&numOfinv);
	}

	printf("%d\n",numOfinv);

	terminate(root);

	return 0;    
}