"""
Solution for Repeated String
https://www.hackerrank.com/challenges/repeated-string/problem?h_l=interview&playlist_slugs%5B%5D=interview-preparation-kit&playlist_slugs%5B%5D=warmup
"""
#!/bin/python3

import math
import os
import random
import re
import sys
from collections import Counter

# Complete the repeatedString function below.
# s = 'aba'
# n = 10 -> 7
# 'aba|aba|aba|a'
#   2 | 2 | 2 |1
def repeatedString(s, n):
    """
    Lilah has a string, , of lowercase English letters that she repeated infinitely many times.

    Given an integer, , find and print the number of letter a's in the first  letters of Lilah's infinite string.

    For example, if the string  and , the substring we consider is , the first  characters of her infinite string. There are  occurrences of a in the substring.

    Function Description

    Complete the repeatedString function in the editor below. It should return an integer representing the number of occurrences of a in the prefix of length  in the infinitely repeating string.

    repeatedString has the following parameter(s):

    s: a string to repeat
    n: the number of characters to consider
    Input Format

    The first line contains a single string, .
    The second line contains an integer, .

    Constraints

    For  of the test cases, .
    Output Format

    Print a single integer denoting the number of letter a's in the first  letters of the infinite string created by repeating  infinitely many times.

    Sample Input 0

    aba
    10
    Sample Output 0

    7
    Explanation 0
    The first  letters of the infinite string are abaabaabaa. Because there are  a's, we print  on a new line.

    Sample Input 1

    a
    1000000000000
    Sample Output 1

    1000000000000
    Explanation 1
    Because all of the first  letters of the infinite string are a, we print  on a new line.

    Args:
        s(str):
        n(int):

    Returns:
        int:

    """
    divider, remainer = divmod(n, len(s))
    a_counter = Counter(s)['a'] * divider
    for i in range(remainer):
        if s[i] == 'a':
            a_counter += 1
    return a_counter

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    s = input()

    n = int(input())

    result = repeatedString(s, n)

    fptr.write(str(result) + '\n')

    fptr.close()
