569_C. Primes or Palindromes?_9693
Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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 (<image>, <image>).

Output

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

Examples

Input

1 1


Output

40


Input

1 42


Output

1


Input

6 4


Output

172
import java.util.Arrays;
import java.util.Scanner;

/**
 * Created by poly on 4/13/15.
 */
public class Main {
    public static void main(String argv[]) {
        Scanner scanner = new Scanner(System.in);
        int max = 2000000;
        int[] primes = new int[max];
        int[] plaindromes = new int[max];
        int sqrt = (int) (Math.sqrt(max)) + 1;
        Arrays.fill(primes, 1);
        primes[0] = 0;
        primes[1] = 0;
        for (int i = 2; i <= sqrt; i++) {
            for (int j = i * 2; j < max; j += i) {
                primes[j] = 0;
            }
        }
        for (int i = 1; i < max; i++) {
            String s = String.valueOf(i);
            boolean plaindrom = true;
            for (int j = 0; j < s.length(); j++) {
                if (s.charAt(j) != s.charAt(s.length() - 1 - j)) {
                    plaindrom = false;
                    break;
                }
            }
            if (plaindrom) {
                plaindromes[i] = 1;
            }
        }
//        System.out.println(Arrays.toString(primes));
//        System.out.println(Arrays.toString(plaindromes));
        int p = scanner.nextInt();
        int q = scanner.nextInt();
        int pi = 0;
        int rub = 0;
        int ans = 0;
        for (int i = 1; i < max; i++) {
            if (primes[i] > 0) pi++;
            if (plaindromes[i] > 0) rub++;
            if (p * rub >= q * pi) {
                ans = Math.max(ans, i);
            }
        }
        System.out.println(ans);
    }
}

6 4
 1 1
 1 42
 238 9996
 9999 9999
 620 35
 9999 9998
 6811 5416
 10000 9999
 42 1
 10000 10000
 241 10000
 5858 674
 1000 10000
 1307 3420
 4 9
 10000 239
 5 8
 16 60
 999 10000
 7 11
 940 480
 214 210
 3 1
 7 267
 239 10000
 106 6
 238 9655
 9999 6698
 987 35
 9854 9998
 11275 5416
 10000 1538
 5858 495
 1854 3420
 7 9
 5 5
 29 60
 12 11
 955 480
 214 248
 3 2
 106 10
 9999 5278
 987 39
 11275 615
 10100 1538
 5858 304
 7 10
 5 2
 1738 480
 145 10
 6343 5278
 11275 351
 11000 1538
 8689 304
 35 4
 48 10000
 999 11000
 7 388
 239 11000
 1 2
 1 45
 238 5291
 10351 9998
 48 10010
 1854 4320
 35 60
 1526 11000
 12 16
 80 248
 8 388
 379 11000
 2 45
 238 2119
 10351 10081
 9 10010
 3498 4320
 9 10
 1 3
 2413 11000
 12 29

172
 40
 1
 1
 40
 251262
 40
 66
 40
 1179858
 40
 1
 71118
 1
 1
 10
 1168638
 16
 1
 1
 16
 1372
 40
 2530
 1
 1
 250300
 1 162 514872 36 1486 11620 146358 12 22 40 10 46 1408 28 172 127296 1296 429508 263862 11940 283026 18 1912 3448 190908 66 652836 15580 526776 100332 1 1 1 1 12 1 1 40 1 1 12 1 22 1 1 1 1 1 40 1 22 28 1 1 1