#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int min(int a, int b) {
    return a < b ? a : b;
}

int max(int a, int b) {
    return a > b ? a : b;
}

int solution(char* s, int k) {
    int n = strlen(s);
    
    // Count total characters
    int total_a = 0, total_b = 0, total_c = 0;
    for (int i = 0; i < n; i++) {
        if (s[i] == 'a') total_a++;
        else if (s[i] == 'b') total_b++;
        else total_c++;
    }
    
    // Check if solution is possible
    if (total_a < k || total_b < k || total_c < k) {
        return -1;
    }
    
    if (k == 0) {
        return 0;
    }
    
    int min_minutes = n;
    
    // Try all possible left counts
    for (int left = 0; left <= n; left++) {
        // Count characters taken from left
        int left_a = 0, left_b = 0, left_c = 0;
        for (int i = 0; i < left; i++) {
            if (s[i] == 'a') left_a++;
            else if (s[i] == 'b') left_b++;
            else left_c++;
        }
        
        // Calculate needed from right
        int need_a = max(0, k - left_a);
        int need_b = max(0, k - left_b);
        int need_c = max(0, k - left_c);
        
        // Try to get needed characters from right
        int right_a = 0, right_b = 0, right_c = 0;
        int right = 0;
        
        while (right < n - left && (right_a < need_a || right_b < need_b || right_c < need_c)) {
            char ch = s[n - 1 - right];
            if (ch == 'a') right_a++;
            else if (ch == 'b') right_b++;
            else right_c++;
            right++;
        }
        
        // Check if we got enough
        if (right_a >= need_a && right_b >= need_b && right_c >= need_c) {
            min_minutes = min(min_minutes, left + right);
        }
    }
    
    return min_minutes;
}

int main() {
    char s[100001];
    int k;
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &k);
    
    int result = solution(s, k);
    printf("%d\n", result);
    
    return 0;
}