#include <iostream>

using namespace std;

const int N = 100010, M = 1000010;

int n, m;
char p[N], s[M];
int ne[N];

int main()
{
    cin >> n >> p + 1 >> m >> s + 1;

    for (int i = 2, j = 0; i <= n; i ++)
    {
        while (j && p[i] != p[j + 1]) j = ne[j];
        //找一个前缀与p[j + 1]的连接等于以p[i]结尾的后缀结尾
        //不断找尽可能长度j = ne[j]知道找到或者j = 0
        if (p[i] == p[j + 1]) j ++; //如果找到就把j++
        ne[i] = j; 
    }

    // kmp匹配
    for (int i = 1, j = 0; i <= m; i ++)
    {
        while (j && s[i] != p[j + 1]) j = ne[j]; // 不等就尽可能找个长一点的前缀与现在后缀匹配
        if (s[i] == p[j + 1]) j ++; //找到或者j为0 此时判断一下能不能j++
        if (j == n)
        {
            printf("%d ", i - n);
            j = ne[j];
        }
    }

    return 0;
}