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>

long long count = 0;


void merge(int A[],int left,int mid,int right) 
{
  int i,j=0,k;
  int n1 = mid-left, n2 = right - mid; 
  int L[n1+1],R[n2+1];
   
  for(i = 0;i < n1;i ++){
    
    L[i] = A[left+i];
  }
  
  for(i = 0;i < n2;i ++){
    R[i] = A[mid+i];
  }
  
  L[n1] = (1e+9)+1;
  R[n2] = (1e+9)+1;

  i = 0;
  
  for(k = left;k < right;k ++)
    {
      if(L[i]<=R[j]){
	A[k] = L[i ++];
      }else {
	A[k] = R[j ++];
	count += n1-i;
      }
    }
}
 
void mergeSort(int A[],int left,int right) 
{
  int mid;
  
  if(left+1 < right) {
    mid = (left+right)/2;
    mergeSort(A,left,mid);
    mergeSort(A,mid,right);
    merge(A,left,mid,right);
  }
}
 
int main() 
{
  int i,n,S[200000];
  
  scanf("%d",&n);
  
  for(i = 0;i < n;i ++) 
    {
      scanf("%d",&S[i]);
    }
  
  mergeSort(S,0,n);
  
  printf("%ld\n",count);
   
  return 0;
}