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: Wrong Answer)

#include <stdio.h>

#define   N   300000
#define  MID  500000
#define INFTY 2000000000

void mergeSort(int*,int,int);
void merge(int*,int,int,int);

unsigned long long count=0;

int main(){

  int i;
  int n;
  int A[N];

  scanf("%d",&n);

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

 return 0;
}

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);
  }
}

void merge(int *A,int left,int mid,int right){

  int i;
  int j;
  int k;
  int n1 = mid - left;
  int n2 = right - mid;
  int L[MID];
  int R[MID];
  
  for(i=0;i<n1;i++){
    L[i] = A[left+i];
  }
  
  for(j=0;j<n2;j++){
    R[j] = A[mid+j];
  }
    L[n1] = INFTY;
    R[n2] = INFTY;
    
    i = 0;
    j = 0;
    
  for(k=left;k<right;k++){
    if(L[i] <= R[j]){
      A[k]=L[i];
      i = i + 1;
    }
    else{ 
      A[k] = R[j];
      j = j + 1;
      if(L[i] !=  INFTY) count += n1 - i;
    }
  }
}