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>
#include<stdlib.h>
#include<math.h>

int main(){
    long long int L,R;
    scanf("%lld %lld",&L,&R);
    if(R-L>3000)printf("0\n");
    //L=L%2019;
    //R=R%2019;
    long long int min=2018;
    for(int i=L;i<R;i++){
        for(int j=L+1;j<R+1;j++){
            if(min > ((i*j)%2019))min=((i*j)%2019);
        }
    }
    printf("%d",min);
 
}