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 9.2.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <tuple>
#include <vector>

using namespace std;
typedef long long ll;

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define reps(i, f, n) for (ll i = (f); i < (n); i++)
#define repr(i, n) for (ll i = n; i >= 0; i--)
#define repv(v) for (auto it = (v).begin(); it != (v).end(); it++)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define pb push_back
#define INIT    \
    cin.tie(0); \
    ios::sync_with_stdio(false);

template <class T>
inline bool chmax(T& a, T b) {
    return a = (a < b) ? b : a;
}
template <class T>
inline bool chmin(T& a, T b) {
    return a = (a > b) ? b : a;
}

ll const INF = 1LL << 60;
ll const MOD = 1000000007;

int main() {
    INIT;

    ll L, R;
    cin >> L >> R;

    if (L / 2019 != R / 2019) {
        cout << 0 << endl;
        return 0;
    }

    ll ans = INF;
    reps(i, L, R + 1) {
        // cout << i * i + 1 << "," << (i * i + 1) % 2019 << endl;
        chmin(ans, (i * i + 1) % 2019);
    }

    cout << ans << endl;

    return 0;
}