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>
#define INFTY 1000000000
int count=0;

void merge(int *a,int left,int mid,int right){
  int n1,n2,l[500000],r[500000];
  int i,j,k;

  n1 = mid-left;
  n2 = right-mid;
  for(i=0;i<n1;i++) l[i] = a[left+i];
  for(i=0;i<n2;i++) r[i] = a[mid+i];
  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++;
    }
    else{
      a[k] = r[j];
      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 n,a[210000];
  int left,right;
  int i;

  scanf("%d",&n);
  for(i=0;i<n;i++) scanf("%d",&a[i]);
  left = 0;
  right = n;

  mergeSort(a,left,right);
  printf("%d\n",count);
  
  return 0;
}