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: Accepted)

#include<stdio.h>

int main(void){
    int l,r,a,b,i,j,ans,min=3000;
    scanf("%d %d",&l,&r);
    a=l;
    b=r;
    l%=2019;
    r%=2019;
    //printf("%d %d ",l,r);
    if(l>=r){
        printf("0");
    }else{
        if(a+2019<b){
            printf("0");
        }else{
            a=a%2019;
            b=b%2019;
            for(i=a;i<b;i++){
                for(j=i+1;j<b+1;j++){
                    ans=(i*j)%2019;
                    if(ans<min){
                        min=ans;
                    }
                }
            }
            printf("%d",min);
        }
    }
    return 0;
}