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)

#include <stdio.h>
int main(){
  int l, r, ans=99999;
  int lst[99999];
  scanf("%d",&l);
  scanf("%d",&r);
  l %= 2019;
  r %= 2019;
  for(int i=l;i<r;i++){
    for(int j=i+1;j<r+1;j++){
      if(ans>(i*j)%2019){
        ans = (i*j)%2019;
        //printf("%d\n",ans);
      }
    }
  }
  printf("%d\n",ans);

}
/*"""
l, r = map(int, input().split())
l %= 2019
r %= 2019

lst = []

for i in range(l,r):
  for j in range(i+1,r+1):
    lst.append(i*j%2019)
print(min(lst))
#*/