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>

#define MaxNum 200000



int main( void )
{
    unsigned long int num=0;
    unsigned long int i,j=0;
    unsigned long int cnt = 0;    // 反転数
    unsigned long int tmp = 0;
    
    unsigned long int sita = 0;
    unsigned long int ue = 0;
    
    unsigned long int  A[MaxNum*2] = {};
    
    scanf("%lu",&num);
    if(num == 0)
    {
        printf("0\n");
        return 0;
    }
    
    sita=num-1;
    ue=num-1;
    
    scanf("%lu", &A[sita]);
    
    for(i=1; i<num; i++)
    {
        scanf("%lu", &tmp);
        
        if(tmp < A[sita])
        {
            sita--;
            A[sita]=tmp;
            cnt=cnt+i;
        }
        else
        {
            for(j=ue+1; j>sita; j--)
            {
                if( tmp < A[j-1] )
                {
                    A[j] = A[j-1];
                    cnt++;
                }
                else
                {
                    break;
                }
            }
            A[j]=tmp;
            ue++;
        }
        

    }
    
    printf("%lu\n", cnt);
    
    return 0;
}