1619： Prime Distance

//#pragma GCC optimize(2)
//#include <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define rush() int T;cin>>T;while(T--)
#define go(a) while(cin>>a)
#define ms(a,b) memset(a,b,sizeof a)
#define E 1e-8
#define debug(a) cout<<"*"<<a<<"*"<<endl
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define PAUSE system("pause")
using namespace std;
typedef long long ll;
typedef unsigned ui;
typedef unsigned long long ull;
typedef pair<int,int> Pair;
const int inf=0x7f7f7f7f;
const int mod=100003;
const int N=1e6+5;

    int n,m,t;
    ll i,j,k;
    bool vis[N];
    ll prime[N],num,pos[N];

void init(){
    for(ll i=2;i<50000;i++)
	  if(!vis[i]){
	  	prime[++num]=i;
	  	for(ll j=i*2;j<50000;j+=i)
          vis[j]=1;
	  }
}
int main()
{
    //IOS;
    init();
    ll R,L;
    while(scanf("%lld%lld",&L,&R)==2){
        ms(vis,0);
        if(L==1) vis[0]=1;//特殊标记，1 不是质数
        //将 [L,R] 区间内的合数标记出来
        for(i=1;i<=num;i++){
            for(j=L/prime[i];j<=R/prime[i];j++){
                if(j>1) vis[prime[i]*j-L]=1;
            }
        }
        int cnt=0;
        for(i=L;i<=R;i++){
            if(!vis[i-L]) pos[++cnt]=i;
        }
        ll maxx=0,minn=2147483647,x,nx,y,ny,dis;
        for(i=1;i<cnt;i++){
            dis=pos[i+1]-pos[i];
            if(dis>maxx){
                maxx=dis;
                x=pos[i+1];
                nx=pos[i];
            }
            if(dis<minn){
                minn=dis;
                y=pos[i+1];
                ny=pos[i];
            }
        }
        if(!maxx) printf("There are no adjacent primes.\n");
        else printf("%lld,%lld are closest, %lld,%lld are most distant.\n",ny,y,nx,x);
    }
    return 0;
}