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: Accepted)

#include <stdio.h>

int main(void){
    int L, R, x, y;

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

    if((R - L) >= 2019){
        printf("0\n");
        return 0;
    }
  
    y = 2018;

    for(int i = L; i < R; i++){
        for(int j = i + 1; j <= R; j++){
            x = (i%2019) * (j%2019) % 2019;
            if(x<y){
                y = x;
            }
        }
    }
    printf("%d\n",y);
    return 0;
}