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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(void){
  long L,R;
  scanf("%ld",&L);
  scanf("%ld",&R);
  long min=2018;
  long l=L%2019;
  long r=R%2019;
    for(long i=l;i<=2019;i++){
      for(long j=i+1;j<=r;j++){
        if( min > ((i%2019)*(j%2019)) % 2019){
          min = ((i%2019)*(j%2019))%2019;
        }
      }
    }
  printf("%ld",min);
  return 0;
}