1 #include<iostream>
2 #include<cstring>
3 using namespace std;
4 const int maxn=1e6+10;
5
6 long long l,r,prime[maxn],vis[maxn],p[maxn],a[maxn];
7 int tot=0;
8
9 int pre(long long n)
10 {
11 for(int i=2;i<=n;i++)
12 {
13 if(!vis[i])
14 {
15 vis[i]=i;
16 prime[++tot]=i;
17 }
18 for(int j=1;j<=tot;j++)
19 {
20 if(i*prime[j]>n)
21 {
22 break;
23 }
24 if(prime[j]>vis[i])
25 {
26 break;
27 }
28 vis[i*prime[j]]=prime[j];
29 }
30 }
31 }
32 int main()
33 {
34 pre(5e4+10);
35 while(cin>>l>>r)
36 {
37 memset(p,0,sizeof(p));
38 if(l==1)
39 {
40 l=2;
41 }
42 long long cnt=0,min1,min2,max1,max2,minn=0x3f3f3f3f,maxx=-0x3f3f3f3f;
43 for(int j=1;j<=tot;j++)
44 {
45 for(int i=l/prime[j];i<=r/prime[j];i++)
46 {
47 while(i<=1)
48 {
49 i++;
50 }
51 if(prime[j]*i<l||prime[j]*i>r)
52 {
53 continue;
54 }
55 p[prime[j]*i-l]=1;
56 }
57 }
58 for(int i=l;i<=r;i++)
59 {
60 if(p[i-l]==0)
61 {
62 a[++cnt]=i;
63 }
64 }
65 if(cnt<=1)
66 {
67 //cout<<cnt<<endl;
68 cout<<"There are no adjacent primes."<<endl;
69 continue;
70 }
71 for(int i=1;i<=cnt-1;i++)
72 {
73 if(a[i+1]-a[i]<minn)
74 {
75 minn=a[i+1]-a[i];
76 min1=a[i+1];
77 min2=a[i];
78 }
79 if(a[i+1]-a[i]>maxx)
80 {
81 maxx=a[i+1]-a[i];
82 max1=a[i+1];
83 max2=a[i];
84 }
85 }
86 //cout<<cnt<<endl;
87 cout<<min2<<","<<min1<<" are closest, "<<max2<<","<<max1<<" are most distant."<<endl;
88 }
89 return 0;
90 }
