//这回只花了114514min就打完了。
//真好。记得多手造几组。
#include <bits/stdc++.h>
//#define int long long
#define pai 3.141592653589793238462643383279502884197169399375105820
#define MOD 998244353
#define eps 0.00000001
inline int min(int a,int b) {return a<b?a:b;}
inline int max(int a,int b) {return a>b?a:b;}
#define ULL unsigned long long
#define LL long long
#define INF 0x3f3f3f3f
#define INF_LL 0x3f3f3f3f3f3f3f3f
static char buf[1000000],*paa=buf,*pd=buf;
static char buf2[1000000],*pp=buf2;
#define getchar() paa==pd&&(pd=(paa=buf)+fread(buf,1,1000000,stdin),paa==pd)?EOF:*paa++
inline void pc(char ch){
	if(pp-buf2==1000000) fwrite(buf2,1,1000000,stdout),pp=buf2;
	*pp++=ch;
}
inline void pcc(){
	fwrite(buf2,1,pp-buf2,stdout);
	pp=buf2;
}
inline int read(void) {
	int w=1;
	register int x(0);register char c(getchar());
	while(c<'0'||c>'9') {if(c=='-') w=-1;c=getchar();}
	while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return w*x;
}
inline LL readl(void) {
	int w=1;
	register LL x(0);register char c(getchar());
	while(c<'0'||c>'9') {if(c=='-') w=-1;c=getchar();}
	while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return w*x;
}
void write(int x) {
	if(x<0) pc('-'),x=-x;
	static int sta[20];
	int top=0;
	do {
		sta[top++]=x%10,x/=10;
	} while(x);
	while(top) pc(sta[--top]+48);
}
void we(int x) {
	write(x);
	pc('\n');
}
inline bool cmp_xi(int a,int b) {return a<b;}
inline bool cmp_da(int a,int b) {return a>b;}
int N,as,dp[2][1<<18],s[1<<18];
LL r[18];
inline int md(int x) {
	return x>=MOD?x-MOD:x;
}
signed main(void) {
    //freopen("m.in","r",stdin);
    //freopen("m.out","w",stdout);
	N=read();
	for(int i=0;i<N;i++) {
		r[i]=readl();
	}
	char c=getchar();
	while(c!='0'&&c!='1') c=getchar();
	for(int i=0;i<(1<<N);i++) {
		s[i]=c-'0';
		c=getchar();
	}
	dp[0][(1<<N)-1]=1;
	bool f=0;
	for(int i=59;i>=0;i--) {
		memset(dp[f^1],0,sizeof(dp[f^1][0])*(1<<N));
		for(int j=0;j<(1<<N);j++) {
			if(!dp[f][j]) continue;
			for(int k=0;k<(1<<N);k++) {
				if(!s[k]) continue;
				int zq=0,fq=0;
				for(int x=0;x<N;x++) {
					if((((j&k)>>x)&1)&&((r[x]>>i)&1)==0) {
						fq=1;
						break;
					}
					if(((r[x]>>i)&1)==((k>>x)&1)&&((j>>x)&1)) zq|=(1<<x);
				}
				if(!fq) dp[f^1][zq]=md(dp[f^1][zq]+dp[f][j]);
			}
		}
		f^=1;
	}
	for(int i=0;i<(1<<N);i++) as=md(as+dp[f][i]);
	printf("%d",as);
    return 0;
}