#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

vector<bool> isprime;
vector<int> primes;
void sieve(int n) {
	if((int)isprime.size() >= n + 1) return;
	isprime.assign(n + 1, true);
	isprime[0] = isprime[1] = false;
	int sqrtn = (int)(sqrt(n * 1.) + .5);
	for(int i = 2; i <= sqrtn; i ++) if(isprime[i]) {
		for(int j = i * i; j <= n; j += i)
			isprime[j] = false;
	}
	primes.clear();
	for(int i = 2; i <= n; i ++) if(isprime[i])
		primes.push_back(i);
}

int main() {
	sieve(200000);
	long long L; long long R;
	while(~scanf("%lld%lld", &L, &R)) {
		pair<int, ll> ans(-1, -1LL);
		for(int p : primes) {
			if((ll)p * p > R) break;
			int q = *(upper_bound(primes.begin(), primes.end(), R / p) - 1);
			if(p <= q && L <= (ll)p * q)
				amax(ans, mp(p, (ll)p * q));
		}
		if(R - L <= 200000) {
			vector<int> minp((int)(R - L + 1), -1);
			for(int p : primes) {
				if((ll)p * p > R) break;
				for(ll t = (L + p - 1) / p * p; t <= R; t += p) {
					int i = (int)(t - L);
					if(minp[i] == -1)
						minp[i] = p;
				}
			}
			rep(i, minp.size()) {
				int p = minp[i];
				if(p != -1)
					amax(ans, mp(p, (ll)L + i));
			}
		}
		printf("%lld\n", ans.second);
	}
	return 0;
}