Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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  true 1;
#define false (-1)
int n,cnt;
  long a[200000];
void bubbleSort(){
    long i,j,key;
    int han;
    for(i=0;i<n;i++){
        han=true;
        for(j=n-1;j>i;j--){
	  if(a[j]<a[j-1]){
            key=a[j-1];
	    a[j-1]=a[j];
	    a[j]=key;
	    cnt++;
            han=false;
	  }
	}
        if(han==1) break;
    }
    return;
}


int main(){

  long i,count;
  
  scanf("%d",&n);
  for(i=0;i<n;i++) scanf("%ld",&a[i]);
  bubbleSort();
    printf("%ld\n",cnt);
 
return 0;
}