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 ,i, j ;
    int isum ,jsum ,iminsum ,jminsum ,minsum;

    iminsum =2019;
    jminsum =2019;

    scanf("%d", &L);
    scanf("%d" ,&R);

    i=L;
    
    for(j= i+1; j <= R ; j++){
        
        jsum = i*j%2019;

        if(jsum < jminsum){
            jminsum = jsum;
        }

        for(i = L;i <= j-1; i++){
            isum = i*j%2019;

            if(isum <iminsum){
                iminsum = isum;
            }

        }

        if(iminsum < jminsum){
            minsum =iminsum;
        }
        else{
            minsum =jminsum;
        }
    }


    printf("%d\n", minsum );
}