CCC '17 J2 - Shifty Sum

Canadian Computing Competition: 2017 Stage 1, Junior #2

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 $12 + 120 + 1\,200 + 12\,000 = 13\,332$.

Input Specification

The first line of input contains the number $N$ $(1 \le N \le 10\,000)$. The second line of input contains $k$, the number of times to shift $N$ $(0 \le k \le 5)$.

Output Specification

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

Sample Input

Sample Output