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)

/*
	submission # - User: herp_sy
	https://atcoder.jp/contests/

	coding: utf-8
	lang: C (GCC 5.4.1)
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <stdbool.h>

#define _CRT_SECURE_NO_WARNINGS
#define TLong long long
#define TBMod 1000000007

int main(int argc, char const *argv[])
{
	TLong l,r;
	TLong min = 2019;
	scanf("%lld%lld", &l,&r);
	if(r - l >= 2019)	puts("0");
	else{
		for (int i = l; i <= r ; ++i)
		{
			for (int j = i + 1; j <= r; ++r)
			{
				min = ((i * j) % 2019 < min) ? (i * j) % 2019 : min;
				if(min == 0){
					puts("0");
					return 0;
				}
			}
		}
		printf("%lld\n", min);
	}
	return 0;
}