Problem:
Score : 300 points

Problem Statement

You are given two non-negative integers L and R.
We will choose two integers i and j such that L \leq i < j \leq R.
Find the minimum possible value of (i \times j) \mbox{ mod } 2019.

Constraints

All values in input are integers.

0 \leq L < R \leq 2 \times 10^9

Input

Input is given from Standard Input in the following format:

L R

Output

Print the minimum possible value of (i \times j) \mbox{ mod } 2019 when i and j are chosen under the given condition.

Sample Input 1

2020 2040

Sample Output 1

2

When (i, j) = (2020, 2021), (i \times j) \mbox{ mod } 2019 = 2.

Sample Input 2

4 5

Sample Output 2

20

We have only one choice: (i, j) = (4, 5).

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int main(){
  int L,R,i,j,min,temp;
  
  scanf("%d %d",&L,&R);
  
  for(i=L; i<=R; i++){
  	if(i%2019==0){min=0;}
  }
  
  L=L%2019;
  R=R%2019;
  min=2018;
  

  for(i=L; i<R; i++){
  	for(j=i+1; j<=R; j++){
      temp=(i*j)%2019;
      if(min>=temp){min=temp;}
    }  
  }
  
  printf("%d\n",min);
  
  return(0);
}