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: Time Limit Exceeded)

#include <stdio.h>

int main(void) {
	int L,R,min = 2019;
	scanf("%d%d",&L,&R);
	while(L>=2019){
		L-=2019;
		R -= 2019;
	}
	for(int i=L; i<R; i++){
		for(int j=i+1; j<=R; j++){
			if((i*j)%2019 <= min){
				min = (i*j)%2019;
			}
		}
	}
	printf("%d\n",min);
	return 0;
}