CF625D.Finals in arithmetic

Vitya is studying in the third grade. During the last math lesson all the pupils wrote on arithmetic quiz. Vitya is a clever boy, so he managed to finish all the tasks pretty fast and Oksana Fillipovna gave him a new one, that is much harder.

Let's denote a flip operation of an integer as follows: number is considered in decimal notation and then reverted. If there are any leading zeroes afterwards, they are thrown away. For example, if we flip 123 the result is the integer 321, but flipping 130 we obtain 31, and by flipping 31 we come to 13.

Oksana Fillipovna picked some number a without leading zeroes, and flipped it to get number a__r. Then she summed a and a__r, and told Vitya the resulting value n. His goal is to find any valid a.

As Oksana Fillipovna picked some small integers as a and a__r, Vitya managed to find the answer pretty fast and became interested in finding some general algorithm to deal with this problem. Now, he wants you to write the program that for given n finds any a without leading zeroes, such that a + a__r = n or determine that such a doesn't exist.

维佳正在上三年级。在最近的一节数学课上，所有学生都完成了一次算术小测验。维佳是个聪明的孩子，因此他很快完成了所有题目，于是奥克萨娜·菲利波夫娜又给了他一道更难的新题。

我们定义一个整数的“翻转”操作如下：将该数以十进制形式写出，然后将其各位数字反转；若反转后出现前导零，则将其全部去掉。例如，翻转 $123$ 得到整数 $321$；翻转 $130$ 得到 $31$；而翻转 $31$ 则得到 $13$。

奥克萨娜·菲利波夫娜选取了一个不含前导零的数 $a$，并将其翻转得到数 $a_r$。接着她计算了 $a + a_r$，并将结果 $n$ 告诉了维佳。维佳的任务是找出任意一个满足条件的 $a$。

由于奥克萨娜·菲利波夫娜所选的 $a$ 和 $a_r$ 都是较小的整数，维佳很快找到了答案，并开始思考解决这一问题的一般性算法。现在，他希望你编写一个程序：对给定的 $n$，找出任意一个不含前导零的整数 $a$，使得 $a + a_r = n$；若不存在这样的 $a$，则判定其不存在。

The first line of the input contains a single integer n (1 ≤ n ≤ 10100 000).

输入的第一行包含一个整数 $n$（$1 ≤ n ≤ 10^{100\,000}$）。

If there is no such positive integer a without leading zeroes that a + a__r = n then print 0. Otherwise, print any valid a. If there are many possible answers, you are allowed to pick any.

如果不存在不含前导零的正整数 $a$，使得 $a + a_r = n$，则输出 0；否则，输出任意一个满足条件的 $a$。若存在多个可能的答案，任选其一即可。

In the first sample 4 = 2 + 2, a = 2 is the only possibility.

In the second sample 11 = 10 + 1, a = 10 — the only valid solution. Note, that a = 01 is incorrect, because a can't have leading zeroes.

It's easy to check that there is no suitable a in the third sample.

In the fourth sample 33 = 30 + 3 = 12 + 21, so there are three possibilities for a: a = 30, a = 12, a = 21. Any of these is considered to be correct answer.

在第一个样例中，$4 = 2 + 2$，唯一可能的 $a$ 是 $a = 2$。

在第二个样例中，$11 = 10 + 1$，唯一有效的解是 $a = 10$。注意，$a = 01$ 是不合法的，因为 $a$ 不能有前导零。

容易验证第三个样例中不存在合适的 $a$。

在第四个样例中，$33 = 30 + 3 = 12 + 21$，因此 $a$ 有三种可能：$a = 30$、$a = 12$、$a = 21$。其中任意一个都被视为正确答案。