//Rabin-Karp
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#define N 2000001 //lungimea maxima a unui sir
#define mod1 2000003
#define mod2 666013

using namespace std;
char a[N], p[N]; //sirul p este cautat in a
int n, m; // n = |p|; m = |a| 
vector<int> sol;// primele 1000 de aparitii

void rk()
{
	int NR = 0; //nr de aparitii a lui p in a
	int pw1_255=1, i, pw2_255=1; // pw1_255 = 255^(n-1) % mod1 iar pw2_255 = 255^(n-1) % mod2
	int hp1=0, h1=0; //hp1 = hashul 1 al sirului p, h1 = hashul 1 al unei subsecvente din a 
	int hp2=0, h2=0; //la fel pt hashul 2
	
	for(i = 1; i < n; ++i)
	//calculam 255^(n-1) % mod 1/2
		pw1_255 *=255, pw1_255 %= mod1,
		pw2_255 *= 255, pw2_255 %= mod2;
	
	for(i = 1; i <= n; ++i)
	//calculam hashurile lui p
		hp1 = hp1 * 255 + p[i], hp1 %= mod1,
		hp2 = hp2 * 255 + p[i], hp2 %= mod2;
	
	for(i = 1; i < n; ++i)
	//calculam hashurile subsecventei a[1..n-1]
		h1 = h1 * 255 + a[i], h1 %= mod1,
		h2 = h2 * 255 + a[i], h2 %= mod2;
	
	for(i = n; i <= m; ++i)
	{
		//adaugam caracterul a[i] la hashuri
		h1 = h1 * 255 + a[i], h1 %= mod1;
		h2 = h2 * 255 + a[i], h2 %= mod2;
		
		//daca cele 4 hashuri coincid (2 cate 2) => am gasit un pattern
		if(h1 == hp1)
			if(h2 == hp2) if(sol.size() < 1000) sol.push_back(i-n), ++NR;
							else ++NR;
		
		//scadem caracterul a[i-n+1] din hashuri
		h1 -= (a[i-n+1] * pw1_255)% mod1;
		h2 -= (a[i-n+1] * pw2_255) % mod2;
		
		while(h1 < 0) h1 += mod1;
		while(h2 < 0) h2 += mod2;
		
	}
	
	printf("%d\n", NR);
	
	
	for(int i = 0; i < min(1000, (int)sol.size()); ++i) printf("%d ", sol[i]);
	printf("\n");
	
	
	
}

int main()
{
	freopen("strmatch.in","r",stdin);
	freopen("strmatch.out","w",stdout);

	gets(p);
	gets(a);
	n = strlen(p);
	m = strlen(a);
	
	int i;
	//mentinem a[1...m] si p[1...n]
	for(i = n; i > 0; --i) p[i] = p[i-1];
	for(i = m ; i > 0;--i) a[i] = a[i-1];
	
	rk();
	return 0;
}