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.

Input

The first line of the input contains a single integer $*n*$ ($1 ≤ *n* ≤ 10^{100 000}$).

Output

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.

Note

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.