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 bubbleSort(int n, int *array);

int main(void)
{
    int i;

    int n;
    scanf("%d", &n);

    int array[MAX];
    for (i = 0; i < n; i++) {
        scanf(" %d", &array[i]);
    }

    int cnt = bubbleSort(n, array);
    printf("%d\n", cnt);

    return 0;
}

void swap(int id1, int id2, int *array);

int bubbleSort(int n, int *array)
{
    int i, j;

    int cnt = 0;
    for (i = 0; i < n; i++) {
        for (j = n - 1; j > i; j--) {
            if (array[j] < array[j - 1]) {
                swap(j, j - 1, array);
                cnt++;
            }
        }
    }

    return cnt;
}

void swap(int id1, int id2, int *array)
{
    int tmp = array[id1];
    array[id1] = array[id2];
    array[id2] = tmp;
}