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    #include <iostream>
    #include <vector>
    #define bsize 100000//
    using namespace std;

    bool prime[100000];//容量兩千萬的表格
    vector<int> primes;
    void findprime(){ //建構質數表
        for (int i=0;i<bsize;i++){//
            prime[i]=true;
        }
        prime[0]=false;
        prime[1]=false;
        for (int i=2; i<bsize; i++){
                if (prime[i]){
                if((long long)i*i<bsize){//
                    for(int m=i*i; m<bsize; m+=i)
                    prime[m] = false;
                }
                primes.push_back(i);//儲存已知的質數
            }
        }
    }
    void cal(long long l, long long u){
        bool isComposite[1000001] = {false};
        for(int i = 0; i<primes.size() && (long long)primes[i] * primes[i] <= u; ++i){
            const int tmp = primes[i]; //利用tmp存取質數
            long long st;//在合數表中，欲刪除的起始值
            if(l%tmp==0)
                st = l;
            else
                st = l - (l%tmp) + tmp;
            for(long long n = st; n<=u; n+=tmp){
                if(n!=tmp)
                    isComposite[n-l] = true;//把合數刪除;題目說U-L不超過10^6
            }
        }
        if(l==1) isComposite[0] = 1;////////////記得要改 1不是質數 
        long long c1, c2, d1, d2;
        c1 = 0, c2 = u, d1 = 0, d2 = 0;
        int lastPrime = 0;
        for(long long n = l; n<=u; n++){
            if(!isComposite[n-l]){
                if(lastPrime!=0){
                    if(c2 - c1 > n-lastPrime){
                        c2 = n;
                        c1 = lastPrime;
                    }
                    if(d2 - d1 < n-lastPrime){
                        d2 = n;
                        d1 = lastPrime;
                    }
                }
                lastPrime = n;
            }
        }
        if (c1 == 0)
            cout << "There are no adjacent primes." << endl;
        else
        	printf("%lld,%lld are closest, %lld,%lld are most distant.\n", c1, c2, d1, d2);
    }

    int main()
    {
        int l,u;
        findprime();//建立質數表
        while(cin>>l>>u){
            cal(l,u);
        }
        return 0;
    }