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 N 200000
#define INFTY 1000000000
double cnt=0;
void merge(int *,int,int,int);
void mergeSort(int *,int,int);
 
void merge(int *A,int left,int mid,int right){
  int i,j,k;
  int x1,x2;
  int L[N],R[N];
 
  x1=mid-left;
  x2=right-mid;
   
  for(i=0;i<x1;i++){
    L[i]=A[left+i];
  }
 
  for(i=0;i<x2;i++){
    R[i]=A[mid+i];
  }
 
  L[x1]=INFTY;
  R[x2]=INFTY;
 
  i=0;
  j=0;
   
  for(k=left;k<right;k++){
    if(L[i]<R[j]){
      A[k]=L[i];
      i++;
    }else{
      A[k]=R[j];
      cnt=cnt+(x1-i);
      j++;
    }
  }
   
}
 
void mergeSort(int *A,int left,int right){
  int middle;
 
  if(left+1<right){
    middle=(left+right+1)/2;
    mergeSort(A,left,middle);
    mergeSort(A,middle,right);
    merge(A,left,middle,right);
  }
}
 
int main()
{
  int i,n,A[N];
 
  scanf("%d",&n);
 
  for(i=0;i<n;i++){
    scanf("%d",&A[i]);
  }
 
  mergeSort(A,0,n);
 
  printf("%.0f\n",cnt);
 
  return 0;
}