/*
                `-/oo+/-   ``
              .oyhhhhhhyo.`od
             +hhhhyyoooos. h/
            +hhyso++oosy- /s
           .yoooossyyo:``-y`
            ..----.` ``.-/+:.`
                   `````..-::/.
                  `..```.-::///`
                 `-.....--::::/:
                `.......--::////:
               `...`....---:::://:
             `......``..--:::::///:`
            `---.......--:::::////+/`
            ----------::::::/::///++:
            ----:---:::::///////////:`
            .----::::::////////////:-`
            `----::::::::::/::::::::-
             `.-----:::::::::::::::-
               ...----:::::::::/:-`
                 `.---::/+osss+:`
                   ``.:://///-.
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <cmath>

using namespace std;

const int N = 2e6;
const int MOD1 = 666013;
const int MOD2 = 269696969;
const int BASE = 67;

char A[5 + N], B[5 + N];
bool sol[5 + N];

int main() {
    freopen("strmatch.in", "r", stdin);
    freopen("strmatch.out", "w", stdout);
    long long lenA, lenB, hashA1, hashA2, hashB1, hashB2, pow1, pow2;
    fgets(A, 3 + N, stdin);
    fgets(B, 3 + N, stdin);
    lenA = strlen(A) - 1;
    lenB = strlen(B) - 1;

    if(lenA > lenB) {
        printf("0\n");
        return 0;
    }

    hashA1 = hashA2 = 0;
    pow1 = pow2 = 1;

    for(int i = 0; i < lenA; i++) {
        hashA1 = (hashA1 * BASE + A[i]) % MOD1;
        hashA2 = (hashA2 * BASE + A[i]) % MOD2;
        if(i > 0) {
            pow1 = (pow1 * BASE) % MOD1;
            pow2 = (pow2 * BASE) % MOD2;
        }
    }

    hashB1 = hashB2 = 0;
    for(int i = 0; i < lenA; i++) {
        hashB1 = (hashB1 * BASE + B[i]) % MOD1;
        hashB2 = (hashB2 * BASE + B[i]) % MOD2;
    }
    int k = 0;

    if(hashA1 == hashB1 && hashA2 == hashB2) {
        sol[0] = true;
        k++;
    }
    for(int i = lenA; i < lenB; i++) {
        hashB1 = ((hashB1 - (B[i - lenA] * pow1) % MOD1 + MOD1) * BASE + B[i]) % MOD1;
        hashB2 = ((hashB2 - (B[i - lenA] * pow2) % MOD2 + MOD2) * BASE + B[i]) % MOD2;

        if(hashA1 == hashB1 && hashA2 == hashB2) {
            k++;
            sol[i - lenA + 1] = true;
        }
    }

    printf("%d\n", k);
    k = 0;
    for(long long i = 0; i < lenB && k < 1000; i++) {
        if(sol[i] == true) {
            k++;
            printf("%lld ", i);
        }
    }
    printf("\n");
    return 0;
}