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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<string>
#include<cmath>
#include<algorithm>/*toupperとかが入っている*/
#include<vector>/*可変長配列vectorのインクルード*/

const int MOD=1000000007;/* 1,000,000,007 で割った余りを出力する　のとき使う*/
using namespace std;/*標準名前空間を利用する。cinやcoutはこれによって利用可能になる*/

long long int L,R;

int main(void) {
  cin >> L >> R;
  
  int max = 2020;
  
  while(L+2019 < R){
    L=L+2019;
  }

  
  for(long long int i=L; i<R; i++){
    for(long long int j=i+1; j<=R; j++){
      
      long long int x = i%2019;
      long long int y = j%2019;
      long long int z = x*y%2019;
      
      if(z<max) max=z;
      
    }
  }
  if(max==2020)max=0;
  
  cout << max << endl;
  return 0;
}