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>
#include <stdlib.h>
#define min(x, y) ((x<y) ? (x) : (y))
#define abs(x)  ((x)>=0 ? (x) : ((x)*-1))

int main(){
	long long int L, R;
	scanf("%lld %lld", &L, &R);

	if (L+(2019-L%2019) < R){
		printf("0");
		return 0;
	}

	
	printf("%lld", (long long int)(L*(L+1)%2019));

	return 0;
}