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: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<cmath>
#include<stack>
#include<queue>
#include<deque>
#include<map>
#include<set>
#include<bitset>
using namespace std;
typedef long long ll;
typedef vector<int> VI;
#define FOR(i,a,n) for(ll (i)=(a);(i)<(n);(i)++)
#define eFOR(i,a,n) for(ll (i)=(a);(i)<=(n);(i)++)
#define SORT(i) sort((i).begin(),(i).end())
#define rSORT(i,a) sort((i).begin(),(i).end(),(a))
constexpr auto INF = 1000000000;
constexpr auto LLINF = 1LL << 62;
constexpr auto mod = 1000000007;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; }return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; }return 0; }

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll l, r;
    cin >> l >> r;

    chmin(r, l + 2019);
    ll ans = INF;
    eFOR(i, l, r)eFOR(j, i + 1, r) {
        chmin(ans, i * j % 2019);
    }
    cout << ans << "\n";
}