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<stdio.h>
int main(){
  int i, j, cal, min = 2018;
  int L, R;
  scanf("%d %d", &L, &R);
    if(R-L>=2019)
{      min=0;
   }
 else
   {    
	for(i = L; i < R; i++){
      for(j = i+1; j <= R; j++){
        cal = (i%2019)*(j%2019)% 2019;
        if(cal < min){
          min = cal;
        }
      }
     }
    }
  printf("%d", min);
}