/*
*POJ 2689 Prime Distance
*给出一个区间[L,U],找出区间内容、相邻的距离最近的两个素数和距离最远的两个素数
*1<=L<U<=2147483647 区间长度不超过1000000、就是要筛选出[L,U]之间的素数
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <stack>
#include <algorithm>
using namespace std;
const int MAXN = 100010;
int prime[MAXN+1];
void getPrime() {
memset(prime, 0, sizeof(prime));
for (int i = 2; i<=MAXN; i++) {
if (!prime[i])
prime[++prime[0]] = i;
for (int j = 1; j<=prime[0] && prime[j] <= MAXN/i; j++) {
prime[prime[j]*i] = 1;
if (i%prime[j] == 0)
break;
}
}
}
bool notprime[1000010];
int prime2[1000010];
void getPrime2(int L, int R) {
memset(notprime, false, sizeof(notprime));
if (L < 2)
L = 2;
for (int i = 1; i<=prime[0]&& (long long)prime[i]*prime[i] <= R; i++) {
int s = L /prime[i] + (L%prime[i] > 0);
if (s == 1)
s = 2;
for (int j = s; (long long)j*prime[i] <= R; j++) {
if ((long long)j*prime[i] >= L)
notprime[j*prime[i]-L] = true;
}
}
prime2[0] = 0;
for (int i = 0; i<=R-L; i++) {
if (!notprime[i])
prime2[++prime2[0]] = i+L;
}
}
int main() {
getPrime();
int L, U;
while (scanf("%d%d",&L,&U) == 2) {
getPrime2(L, U);
if (prime2[0] < 2)
printf("There are no adjacent primes.\n");
else {
int x1 = 0, x2 = 100000000, y1 = 0, y2 = 0;
for (int i = 1; i<prime2[0]; i++) {
if (prime2[i+1]-prime2[i] < x2-x1) {
x1 = prime2[i];
x2 = prime2[i+1];
}
if (prime2[i+1] - prime2[i] > y2-y1) {
y1 = prime2[i];
y2 = prime2[i+1];
}
}
printf("%d,%d are closest, %d,%d are most distant.\n",x1,x2,y1,y2);
}
}
return 0;
}