A. Primes or Palindromes?

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!

Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.

Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.

One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(n) — the number of primes no larger than n, rub(n) — the number of palindromic numbers no larger than n. Rikhail wants to prove that there are a lot more primes than palindromic ones.

He asked you to solve the following problem: for a given value of the coefficient A find the maximum n, such that π(n) ≤ A·rub(n).

Input

The input consists of two positive integers p, q, the numerator and denominator of the fraction that is the value of A (, ).

Output

If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).

Sample test(s)

input

1 1

output

40

input

1 42

output

1

input

6 4

output

172

能够发现不可能无解，极限情况n不大

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              using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i
              
               =0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}bool is_prime(int x){ if (x==1) return 0; Fork(i,2,sqrt(x)) { if (x%i==0) return 0; } return 1;}const int MAXN =10000000;int P[MAXN],siz=0,b[MAXN]={0};void make_prime(int n){ Fork(i,2,n) { if (!b[i]) { P[++siz]=i; } For(j,siz) { if (P[j]*i>n) break; b[P[j]*i]=1; if (i%P[j]==0) break; } }}bool is_pal(int x){ char s[10]; sprintf(s,"%d",x); int p=0,q=strlen(s)-1; while(p
               
                -1;j++,p--) s[j]=s[p]; int x; sscanf(s,"%d",&x); if (x<=n) B[x]=1; for(int j=m;j<=2*m-1;j++) s[j]=s[j+1]; sscanf(s,"%d",&x); if (x<=n) B[x]=1; } }int main(){// freopen("A.in","r",stdin);// freopen(".out","w",stdout); int p,q; cin>>p>>q; make_prime(MAXN-1); make_pal(MAXN-1); int x1=0,x2=0,n=MAXN-1,ans=1,t=1; For(i,n) { if (i==P[t]) x1++,t++; if (B[i]) x2++; if ((ll)(x1)*q<=(ll)(x2)*p) ans=i; } cout<
                
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