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: Time Limit Exceeded)

#include<stdio.h>
#include<math.h>

int main(void){

  long long int L,R,i,j,k;
  scanf("%lld",&L);
  scanf("%lld",&R);
  long long int tmp,min=2019;

  for(i=L;i<R;i++){
    for(j=i+1;j<R+1;j++)
    if(i<j){
     tmp = (i*j) % 2019;
     if(min > tmp){
       min = tmp;
     }
    }
  }

  printf("%lld",min);

  return 0;
}