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<stdlib.h>


int main()
{
	long int L, R;
	int l, k, a, b, i, j, c;
	scanf("%d", &L);
	scanf("%d", &R);

	l = L % 2019;

	k = R - L;

	if (k > 2017)
	{
		printf("0");
		return 0;
	}

	c = l * (l + 1) % 2019;

	for (i = l; i < l + k; i++) 
	{
		for (j = i + 1; j < l + k + 1; j++) 
		{	
			if ( i * j % 2019 < c) 
			{ 
				c = i * j % 2019;
			}
		}

		if (c == 0) 
		{
			break;
		}
	}
	printf("%d", c);
	return 0;
}