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<math.h>
int main(){
    int L,R;
    scanf("%d %d",&L,&R);
    int i,j;
    int k,l;
    int min=2018;
    //int minj=2018;
    int J;
    if(R-L>=2018){
        printf("0\n");
    }
    else{
    for(k=L;k<=R;k++){
        for(l=k+1;l<=R;l++){
            J = (k*l)%2019;
            if(J<min){
                min=J;
            }
        }
        
    }
   // printf("%d %d\n",mini,minj);
    printf("%d\n",min);
    }
    return 0;
}