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 <bits/stdc++.h>
using namespace std;
 
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
typedef pair<int,ii> iii;
typedef pair<int, iii> iiii;
typedef pair<double, ii> dii;
typedef long long ll;
typedef pair<ll, ii> lii;
const int INF = 1000000000;
const double EPS = 1e-9;
#define LSOne(S) (S & (-S))
int mods(int a, int b) { return (b + (a%b)) % b; }


int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    int L, R, n, ans = INF;
    bitset<2025> visited;

    cin >> L >> R;

    int mx = -INF;
    for (int i = L; i <= R; i++) {
        n = i % 2019;
        if (visited[n]) break;
        else visited[n] = 1;
        mx = max(mx, n);
    }


    for (int i = 0; i <= mx; i++) {
        if (!visited[i]) continue;
        for (int j = i+1; j <= mx; j++) {
            if (!visited[j]) continue;
            ans = min(ans, (i*j) % 2019);
        }
    }
    
    cout << ans << '\n';

    return 0;
}