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>
#include <stdlib.h>
#include <math.h>

#define INFTY 1000000000
void rge(int*,int,int,int);
void rgesort(int*,int,int);
int count=0;
int main(){

  int i,n,left,right;

  scanf("%d",&n);
  int A[n];
  left = 0;
  right = n;
  for(i=0;i<n;i++){
    scanf("%d",&A[i]);
  }
  rgesort(A,left,right);

  printf("%d\n",count);
  
  return 0;
  
}

void rge(int *A,int left,int mid,int right){
  int n1,n2,i,j,k,t=0;
  n1 = mid - left;
  n2 = right - mid;
  int L[n1];
  int R[n2];
  for(i=0;i<=n1-1;i++){
    L[i] = A[left+i];
  }
  for(i=0;i<=n2-1;i++){
    R[i] = A[mid+i];
  }
  L[n1] = INFTY;
  R[n2] = INFTY;
  i=0;
  j=0;
  for(k=left;k<=right-1;k++){
    if(L[i]<= R[j]){
      count += t;
      A[k] = L[i];
      i=i+1;
    }
    else{
      t++;
      A[k]=R[j];
      j=j+1;
    }
  }
}
void rgesort(int *A,int left,int right){
  int mid;
  if (left+1 < right){
    mid = (left + right)/2;
    rgesort(A, left, mid);
    rgesort(A, mid, right);
    rge(A, left, mid, right);
  }
}