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 N 200000
int bubblesort(int*);
int n;
int A[N];
int main(){
  int a,i;

  scanf("%d",&n);

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

  a = bubblesort(A);

  printf("%d\n",a);
  
  return 0;
  
}

int bubblesort(int *A){
  int count=0;
  int i,j,hozon;
  for(i=0;i<n;i++){
    for(j=n-1;j>=i+1;j--){
      if(A[j]<A[j-1]){
	hozon = A[j];
	A[j] = A[j-1];
	A[j-1] = hozon;
	count++;
      }
    }
  }
  return count;
}