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;
  printf("L");
  scanf("%d",&L);
  printf("R");
  scanf("%d",&R);
  int max=0;
  if(L<0||R<0){
    printf("error");
  } else {
    for(int i=L;i<=R-1;i++){
      for(int j=i+1;j<=R;j++){
        if( max < (i*j) % 2019){
          max = (i*j)%2019;
        }
      }
    }
  }
    printf("%d\n",max);
}