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: Accepted)

#include<stdio.h>
 
#define M 200000
#define S 2000000000
 
int L[M/2+2],R[M/2+2];
long b1,b2,b3;
 
long mer(int *a,int n,int left,int right,int man){
 
  int i,j,k=0;
  long cou=0;
 
  int na=man-left;
  int nb=right-man;
 
  for(i=0;i<na;i++) L[i]=a[left+i];
  for(i=0;i<nb;i++) R[i]=a[man+i];
 
  L[na]=R[nb]=S;
 
  i=0;
  j=0;
  for(k=left;k<right;k++){
    if(L[i]<=R[j]){
      a[k]=L[i++];
    }else{
       
      a[k]=R[j++];
      cou+=na-i;
    }
  }
  return cou;
}
 
long sort(int *a,int n,int right,int left){
  int man;
  long b1,b2,b3;
   
 
  if(left+1<right){
    man=(left+right)/2;
    b1=sort(a,n,man,left);
    b2=sort(a,n,right,man);
    b3=mer(a,n,left,right,man);
    return b1+b2+b3;
  }else return 0;
}
 
int main(){
  int a[M],n,i;
  long ans=0;
 
  scanf("%d",&n);
 
  for(i=0;i<n;i++){
    scanf("%d",&a[i]);
  }
 
 
  ans=sort(a,n,n,0);
  printf("%ld\n",ans);
 
  return 0;
}