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 (Clang 3.8.0) (judge verdict: Wrong Answer)

    //AtCoder Beginner Contest 133 目指せ3冠、勝ち取れ内定(⊙ꇴ⊙)

#include <stdio.h>
int main(void){
    //変数宣言ここから٩(ˊᗜˋ*)و
    
    char s[100001];
    int n,i,j,count;
    int l,r;
    int min = 2147483647;
    int mode = 0;
    
    //入力処理ここから٩(ˊᗜˋ*)و
    
    scanf("%d %d",&l,&r);
    
    //処理部分ここから٩(ˊᗜˋ*)و
    
    if (l % 2019 == 0){
        mode = 1;
        min = 0;
    }
    
    if (mode == 0){
        for (i = l; i <= r - 1; i++){
            if (i % 2019 == 0){
                min = 0;
                break;
            }
            for (j = l + 1; j <= r; j++){
                count = ((i * j) % 2019);
                if (min > count){
                    min = count;
                    if (min == 0){
                        break;
                    }
                }
                if (j % 2019 == 0){
                    min = 0;
                    break;
                }
            }
            if (min == 0){
                break;
            }
        }
    }
        
    //出力処理ここから٩(ˊᗜˋ*)و
    
    printf("%d",min);
    
}