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 INFINITY 1000000000

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

int count=0;

int main(){

  int i;
  int n,left,right;

  scanf("%d",&n);
  int N[n];
  for(i=0;i<n;i++){
    scanf("%d",&N[i]);
  }
  left=0;
  right=n;
  mergesort(N,left,right);
  printf("%d\n",count);
}
void merge(int *N,int left,int mid,int right){

  int i,j,k;
  int r,l;
    
  l=mid-left;
  r=right-mid;
  int L[l],R[r];
  for(i=0;i<l;i++){
    L[i]=N[left+i];
  }
  for(i=0;i<r;i++){
    R[i]=N[mid+i];
  }
  L[l]=INFINITY;
  R[r]=INFINITY;
  
  i=0;
  j=0;
  for(k=left;k<right;k++){
    if(L[i]<R[j]){
      N[k]=L[i];
      i++;
    }
    else{
      N[k]=R[j];
      j++;
      count+=l-i;
    }
  }
}
void mergesort(int *N,int left,int right){

  int mid,i;
  if(left+1<right){
    mid=(left+right)/2;
    mergesort(N,left,mid);
    mergesort(N,mid,right);
    merge(N,left,mid,right);
  }
}