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 <iostream>
#include <string> 
#include <vector> 
#include <algorithm> 
#include <utility> 
#include <tuple> 
#include <cstdint> 
#include <cstdio> 
#include <map> 
#include <queue> 
#include <set> 
#include <stack> 
#include <deque> 
#include <unordered_map> 
#include <unordered_set> 
#include <bitset> 
#include <cctype> 
#include<math.h>
using namespace std;

#define int long long
//#define endl "\n"
#define fi first
#define se second
#define fr front
#define m_p make_pair
#define mod 1000000007
#define all(v)  v.begin(),v.end()
#define rep(i,n)  for(int i=0; i<(int) (n); i++)
#define all(v)  v.begin(),v.end()
#define vecin(v) for(int i=0; i<(int)v.size(); i++)cin>>v[i];
using namespace std;

int jousu(int, int);
int jousu(int x00, int y00) {
	int z00 = 1;
	for (int i = 0; i < y00; i++) {
		z00 *= x00;
	}
	return z00;
}
int keta(int x00) {
	int z00 = x00;
	int w00 = 0;
	while (z00 != 0) {
		z00 /= 10;
		w00++;
	}
	return w00;
}
int modinv(int a, int m) {
	int b = m, u = 1, v = 0;
	while (b) {
		int t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}




signed main() {
	int l, r, ans=2018;
	cin >> l >> r;
	l %= 2019;
	r %= 2019;
	for (int i = l; i <= r; i++) {
		for (int j = i+1; j <= r; j++) {
			ans = min(ans, (i * j) % 2019);
		}
	}
	cout << ans << endl;
}