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: Wrong Answer)

#include <stdio.h>
#define N 2019

int main(void){
  unsigned long int l,r,ans;
  int count;
  
  scanf("%ld %ld",&l,&r);
  
  ans = (l * (l + 1)) % N;
  count = 0;
  for(unsigned long int i = l; i < r; i++){
    for(unsigned long int j = i + 1; j <= r; j++){
      if(count >= (N - 1)) goto loopend;
      if(ans > (i * j) % N) ans = (i * j) % N;
      count++;
    }
  }
  loopend:
    printf("%ld",ans);
    return 0;
}