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 "pch.h"
#include "stdio.h"
#include "stdlib.h"
#include "math.h"



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

	int nDiff = R - L;
	L = L % 2019;
	if (nDiff > 2019)
		R = L + 2019;
	else
		R = L + nDiff;

	int nMod = (L * (L + 1)) % 2019;
	for (int i = L; i < R; i++)
	{
		for (int j = i + 1; j <= R; j++)
		if (nMod > (i * j) % 2019)
			nMod = (i * j) % 2019;
	}

	printf("%d", nMod);

	return 0;
}