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 PHP (5.6.30) (judge verdict: Wrong Answer)

<?php

const INPUT_DELIMITER = " ";
$input = fgets(STDIN);
list($L, $R) = array_map('intval', explode(INPUT_DELIMITER, $input));


const MOD = 2019;

// 最大がMOD以下の時
if ($R < MOD) {
    echo $L * $R;
    return;
}

// $Lか$Rが2019の倍数を取りうる時
//（差が2019以上であればどっちかが2019の倍数に出来る）
if ($R - $L > MOD) {
    echo 0;
    return;
}

// それ以外のときは地道に計算してみる
$A = MOD;
$B = MOD;
for ($i = $L; $i <= $R; $i++) {
    $mod = $i % MOD;
    if ($A > $mod) {
        $A = $mod;
        continue;
    }
    if ($B > $mod) {
        $B = $mod;
    }
}

echo ($A * $B) % MOD;