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 <string.h>
#include <math.h>

int main()
{
    long l,r,c,min=2019,tmp;
    long i,j;

    scanf("%ld %ld",&l,&r);
    c=r-l;
    if(c>=2019){
        printf("0\n");
        return 0;
    }   
    l = l % 2019;
    r = r % 2019;
    for(i=l;i<=r;i++){
        tmp = ((i % 2019) * (j % 2019)) % 2019;
        if(min > tmp){
            min = tmp;
        }
    }
    
    printf("%ld\n",min);
    

    return 0;
}