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>
 
double count=0;
 
int S[200000],L[100001],R[100001];

void margesort(int *S,int left,int right){
  int mid;
  if(left+1 < right){ 
    mid = (left + right)/2;
    margesort(S,left,mid);
    margesort(S,mid,right);
    marge(S,left,mid,right);
  }
}
void marge(int *S,int left,int mid,int right){
  int n1,n2;
  int i,j=0,k;
 
  n1 = mid - left;
  n2 = right - mid;
 
  for(i=0;i<n1;i++) L[i] = S[left+i];
  L[n1]=-1;
  for(i=0;i<n2;i++) R[i] = S[mid+i];
  R[n2]=-1;
 
  i=0;
  for(k=left;k<right;k++){
  if(L[i] > R[j]){  
    S[k] = L[i];
    i++;
    count += n2 - j;
  }
  else{
    S[k] = R[j];
    j++;
  }
  }
} 
int main(){
  FILE* p;
  int n,i;
  
  scanf("%d",&n);
  for(i=0;i<n;i++)
     scanf("%d",&S[i]);
   
   margesort(S,0,n);
   printf("%.0f\n",count); 
 
   return 0;
}