Problem J2: Shifty Sum

Time limit: 1 second

Problem DescriptionCopy Anchor Link

Suppose we have a number like 12. Let’s define shifting a number to mean adding a zero at the end. For example, if we shift that number once, we get the number 120. If we shift the number again we get the number 1200. We can shift the number as many times as we want.

In this problem you will be calculating a shifty sum, which is the sum of a number and the numbers we get by shifting. Specifically, you will be given the starting number $N$ and a non-negative integer $k$. You must add together $N$ and all the numbers you get by shifting a total of $k$ times.

For example, the shifty sum when $N$ is 12 and $k$ is 1 is: $12 + 120 = 132$. As another example, the shifty sum when $N$ is 12 and $k$ is 3 is $12 + 120 + 1200 + 12000 = 13332$.

Input SpecificationCopy Anchor Link

The first line of input contains the number $N$ $(1 \leq N \leq 10000$). The second line of input contains $k$, the number of times to shift $N$ $(0 \leq k \leq 5)$.

Output SpecificationCopy Anchor Link

Output the integer which is the shifty sum of $N$ by $k$.

Sample InputCopy Anchor Link

12
3

Output for Sample InputCopy Anchor Link

13332