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

Output Specification

Output the integer which is the shifty sum of \(N\) by \(k\) .

Sample Input

12
3

Sample Output

13332