CF568A.Primes or Palindromes?

Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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).

里哈伊尔·穆宾奇克认为，当前素数的定义已经过时，因为素数过于复杂且难以预测。而回文数则另当别论：它具有美学上的吸引力，并拥有一系列显著的性质。请帮助里哈伊尔向科学界证明这一点！

我们先回顾一下：一个数被称为素数，当且仅当它是大于 1 的整数，且除了 1 和它自身之外，不能被任何其他正整数整除。

里哈伊尔将一个数称为回文数，当且仅当它是正整数，且其十进制表示（不含前导零）是一个回文串，即从左到右读与从右到左读完全相同。

素数的一个问题是：它们的数量实在太多了。为此，我们引入如下记号：

• $\pi(n)$ 表示不超过 $n$ 的素数的个数；
• $\mathrm{rub}(n)$ 表示不超过 $n$ 的回文数的个数。

里哈伊尔希望证明：素数的数量远多于回文数的数量。

他请你解决如下问题：给定系数 $A$，求最大的 $n$，使得

$$\pi(n) \leq A \cdot \mathrm{rub}(n).$$

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

输入包含两个正整数 $p$、$q$，分别表示分数 $A$ 的分子与分母（，）。

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

如果这样的最大数存在，则输出它；否则，输出 “Palindromic tree is better than splay tree”（不带引号）。