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(void){
  int l,r,i,j,min=2020,mod,flag=0;
  scanf("%d %d",&l,&r);

  /*if(r-l>=2019){
    min=0;
  }
  else{*/
  for(i=l;i<r;i++){
    for(j=i+1;j<=r;j++){
      if(((i*j)%2019)<min){
        min=(i*j)%2019;
      }
      if(min==0){
        flag=1;
      }
    }
    if(flag==1){
      break;
    }
  }
  //}
  printf("%d",min);
  return 0;
}