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 *A,int n){
  int cont=0;
  int flag=1;
  int i,tmp;
  while(flag==1){
    flag=0;
    for(i=n-1;i>=1;i--){
      if(A[i]<A[i-1]){
	tmp=A[i];
	A[i]=A[i-1];
	A[i-1]=tmp;
	flag=1;
	cont++;
      }
    }
  }
  return cont;
}
int main(){
  int n,i;
  int A[MAX];
  int count; 
  scanf("%d",&n);
  for(i=0;i<n;i++) scanf("%d",&A[i]);
  count = BubbleSort(A,n);
  printf("%d\n",count);
  return 0;
}