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>
#define MOD 2019
#define NUM


int main(void)
{
    
    long long int a,i,j,L,R,I,J;
    scanf("%lld%lld",&L,&R);
    a=(L%MOD)*((L+1)%MOD);
    if(R-L==2019){
        a=0;
    }
    
    for(i=L%MOD;i<=(R-1)%MOD;i++){
        I=i%MOD;
        for(j=(i+1)%MOD;j<=R%MOD;j++){
            J=j%MOD;
            if(a>I*J)
            a=I*J;
        }
    }
   
    printf("%lld\n",a);
    return 0;
}