#include<bits/stdc++.h>
#define ll long long
using namespace std;
void rd(){}
template<typename T,typename... U> void rd(T &x,U&... arg){
	x=0;int f=1,c=getchar();
	while(c<'0'||c>'9'){if(c=='-') f=-1;c=getchar();}
	while(c>='0'&&c<='9') x=x*10+c-48,c=getchar();
	x*=f;rd(arg...);
}
void rds(char *s){
	int c=getchar();
	while(c<'a'||c>'z') c=getchar();
	while(c>='a'&&c<='z') *s=(char)c,s++,c=getchar();
	*s='\0';
}
/*枚举border的长度 分成两种
(1)本来就是S的border 容易贡献到答案中
(2)只需要修改一个字符就变成border 贡献到对应位置
对(2)的判断可以哈希 也可以exkmp*/
inline bool chkmax(int &x,int y){return x<y&&(x=y,1);}
const int maxn=2e6+5,P=331,mod1=998244353,mod2=1e9+7;
const char C='a';
ll h1[maxn],pw1[maxn],h2[maxn],pw2[maxn];
inline ll geth1(int l,int r){return (h1[r]-h1[l-1]*pw1[r-l+1]%mod1+mod1)%mod1;}
inline ll geth2(int l,int r){return (h2[r]-h2[l-1]*pw2[r-l+1]%mod2+mod2)%mod2;}
bool check(int l1,int r1,int l2,int r2){return geth1(l1,r1)==geth1(l2,r2)&&geth2(l1,r1)==geth2(l2,r2);}
int N,z1[maxn],z2[maxn],a[maxn],n,border,ok[maxn][26];
char s[maxn],t[maxn];
void Z(char *s,int *z){
	int n=z[1]=(int)strlen(s+1);
	while(2+z[2]<=n&&s[1+z[2]]==s[2+z[2]]) z[2]++;
	for(int i=3,l=2,r=z[2]+1;i<=n;i++){
		z[i]=max(0,min(r-i+1,z[i-l+1]));
		while(i+z[i]<=n&&s[1+z[i]]==s[i+z[i]]) z[i]++;
		if(i+z[i]-1>r) l=i,r=i+z[i]-1;
	}
}
int main(){
	rds(s+1),rds(t+1),N=(int)strlen(s+1);
	reverse(s+1,s+N+1),Z(s,z2),reverse(z2+1,z2+N+1);
	reverse(s+1,s+N+1),Z(s,z1);
	pw1[0]=pw2[0]=1;
	for(int i=1;i<=N;i++)
		pw1[i]=pw1[i-1]*P%mod1,h1[i]=(h1[i-1]*P+s[i])%mod1,
		pw2[i]=pw2[i-1]*P%mod2,h2[i]=(h2[i-1]*P+s[i])%mod2;
	for(int i=1;i<N;i++){
		int x=z1[N-i+1],y=z2[i];
		if(x==i){chkmax(border,a[++n]=i);continue;}
		if(x+y==i-1)
			chkmax(ok[x+1][s[N-y]-C],i),
			chkmax(ok[N-y][s[x+1]-C],i);
		else if(i-y==N-i+1+x&&check(x+2,i-y-1,i-y+1,N-y-1)&&s[x+1]==s[N-y])
			chkmax(ok[i-y][s[x+1]-C],i);
	}
	for(int i=1,j=0;i<=N;i++){
		if(s[i]==t[i]){printf("%d\n",border);continue;}
		while(j<n&&a[j+1]<i) j++;
		while(j&&a[j]>N-i) j--;
		int Max=max(a[j],ok[i][t[i]-C]);
		printf("%d\n",Max);
	}
	return 0;
}