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>

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

    char buf[256];
    fgets(buf, 256, stdin);
    sscanf(buf, "%d %d", &l, &r);

    if((r -l)/673 >= 1){ //2019 = 673 / 3
        // printf("673 must included.\n");
        // printf("%d", 0);
    }else{
        for (i = l; i <= r-1; i++){
            for (j = i+1; j <= r; j++){
                ijmod = ((long)i * j) % 2019;
                if (min > ijmod){
                    // printf("i = %d j = %d : change %d -> %ld\n", i, j ,min, ijmod);
                    min = ijmod;
                }
            }
        }
        // printf("%d", min);
    }
    printf("%d", min);

}