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 <bits/stdc++.h>
using namespace std;;
#define ll long long
#define rep(i, n) for(int i = 0;i < n;i++)
#define repr(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define INF 1<<30
#define LINF 1LL<<62
#define all(x) (x).begin(), (x).end()
#define mp make_pair
#define pb push_back
const int MOD = 1000000007;
 
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  int l,r; cin >>  l >> r;
  int ans = 2018;
  FOR(i,l,l+2022){
      FOR(j,i+1,i+2022){
          if(l<=i && i <= r && l <= j && j <= r){
              ans = min(ans , (i*j)%2019);
          }
      }
  }
  cout << ans << endl;
}