#include <bits/stdc++.h>
using namespace std;
ifstream f("strmatch.in");
ofstream g("strmatch.out");
string pat, txt;
vector<int> rez;
void computeLPSArray(int M, int *lps);

// Prints occurrences of txt[] in pat[]
void KMPSearch()
{
    int M = pat.size();
    int N = txt.size();

    // create lps[] that will hold the longest prefix suffix
    // values for pattern
    int lps[M];

    // Preprocess the pattern (calculate lps[] array)
    computeLPSArray(M, lps);

    int i = 0; // index for txt[]
    int j = 0; // index for pat[]
    while (i < N)
    {
        if (pat[j] == txt[i])
        {
            j++;
            i++;
        }

        if (j == M)
        {
            rez.push_back(i - j);
            j = lps[j - 1];
        }

        // mismatch after j matches
        else if (i < N && pat[j] != txt[i])
        {
            // Do not match lps[0..lps[j-1]] characters,
            // they will match anyway
            if (j != 0)
                j = lps[j - 1];
            else
                i = i + 1;
        }
    }
}

// Fills lps[] for given patttern pat[0..M-1]
void computeLPSArray(int M, int *lps)
{
    // length of the previous longest prefix suffix
    int len = 0;

    lps[0] = 0; // lps[0] is always 0

    // the loop calculates lps[i] for i = 1 to M-1
    int i = 1;
    while (i < M)
    {
        if (pat[i] == pat[len])
        {
            len++;
            lps[i] = len;
            i++;
        }
        else // (pat[i] != pat[len])
        {
            // This is tricky. Consider the example.
            // AAACAAAA and i = 7. The idea is similar
            // to search step.
            if (len != 0)
            {
                len = lps[len - 1];

                // Also, note that we do not increment
                // i here
            }
            else // if (len == 0)
            {
                lps[i] = 0;
                i++;
            }
        }
    }
}

// Driver program to test above function
int main()
{
    f >> pat >> txt;
    KMPSearch();
    g << rez.size() << '\n';
    for (int i = 0; i < min(int(rez.size()), 1000); i++)
        g << rez[i] << ' ';
    return 0;
}