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 <string.h>

int InSort(int A[], int n)
{
  int sum=0;
    for (int i = 1; i < n; i++)         
    {
        int temp = A[i];                 
        int j = i - 1;                  
        while (j >= 0 && A[j] > temp)    
        {
            A[j + 1] = A[j];sum++;          
            j--;                        /////////   J 位置理解!!!!!!!!!!!!!!!!!!!!! 
        }
        A[j + 1] = temp; // fang jin qu
    }
    return sum;
}
int main ()
{
	int a[200010];
	int t;
	scanf("%d",&t);
	for(int i=0;i<t;i++)
	scanf("%d",&a[i]);
	printf("%d\n",InSort(a,t));
	return 0;
}