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 MIN(int a, int b){return a>b?b:a;}

typedef long long int ll;

int main(void){
    int l,r;
    scanf("%d%d", &l,&r);

    int time;
    if(r-l+1 >= 2019){
        time = 2019;
    }else{
        time = r-l+1;
    }

    int ans = 2019;
    for(int i=0; i < time; i++){
        ans = MIN(ans,(l*(l+i+1))%2019);
    }

    printf("%d\n", ans);
    return 0;
}