//varianta cu un sg hash...

#include <stdio.h>
#include <string.h>
#define NMAX 2000004
#define MOD1 666013
#define MOD2 100007
#define B 73
char nna[NMAX],nnb[NMAX];
int na,nb;
int rez[1005],indice=0;
int main(){  
   int i;
   FILE *fin=fopen("strmatch.in","r");
   FILE *fout=fopen("strmatch.out","w");
   fscanf(fin,"%s%s",nnb,nna);
//   printf("sirul pattern este: %s\n",b);
   na=strlen(nna);
   nb=strlen(nnb);
   //transform a[] si b[]
//   printf("lungimea sirului pattern este %d\n",nb); 
   // printf("\n");
   //practic fiecare nna[] si nnb[] sunt vectori de cifre in baza 26
    long long nr1=0,nr2=0,hashb1=0,hashb2=0;
   //calculez hash pattern
   for(i=0;i<nb;i++){
      hashb1=(hashb1*B+nnb[i])%MOD1;
      hashb2=(hashb2*B+nnb[i])%MOD2;
   }
//printf("hashul patternului este %lld\n",hashb1);

   for(i=0;i<nb;i++){
      nr1=(nr1*B+nna[i])%MOD1;
      nr2=(nr2*B+nna[i])%MOD2;
 }
   //printf("%lld ",nr);
   long long putere1=1,putere2=1;//calc 26^(nb-1)
      for(i=1;i<nb;i++){
        putere1=(putere1*B)%MOD1;
        putere2=(putere2*B)%MOD2;	
      }
  //printf("puterea este %lld\n",putere);
   if((nr1==hashb1)&&(nr2==hashb2) ){
      rez[indice++]=0;
      //printf("0\n");
   }
   for(i=1;i<=(na-nb);i++){
     // fprintf(fout,"%lld %lld\n",nr1,nr2);
      nr1=(nr1-(putere1*nna[i-1])%MOD1+MOD1)%MOD1;
      nr1=(nr1*B+nna[i+nb-1])%MOD1; 
 
      nr2=(nr2-(putere2*nna[i-1])%MOD2+MOD2)%MOD2;
      nr2=(nr2*B+nna[i+nb-1])%MOD2; 

    //printf("%lld ",nr);
      if((nr1==hashb1)&&(nr2==hashb2)){
        rez[indice++]=i;
        if(indice==1000)break;
       //printf("%d\n",i);
      }
   }

    fprintf(fout,"%d\n",indice);
    for(i=0;i<indice;i++)
      fprintf(fout,"%d ",rez[i]);   
   //printf("%s %s\n",a,b);
return 0;
}