#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define fi first
#define sc second
#define mkp make_pair
#define pii pair<int,int>
typedef long long ll;
const int N=10005,oo=1e9,mod=1e9+7;
inline int read() {
    int x=0,flag=0;char ch=getchar();
    while(ch<'0'||ch>'9') {flag|=(ch=='-');ch=getchar();}
    while('0'<=ch&&ch<='9') {x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return flag?-x:x;
}
inline int mx(int x,int y) {return x>y?x:y;}
inline int mn(int x,int y) {return x<y?x:y;}
inline void swp(int &x,int &y) {x^=y^=x^=y;}
inline int as(int x) {return x>0?x:-x;}

int n,r1,r2,r3,A,B,C,D,Cs[N][N],Sm[N][N],psum[N];
char s1[N],s2[N],s3[N];
#define S(i,j) Sm[(j)+D][i]
inline int calc(int n,int m) {
    if(n<m||m<0) return 0;
    else return Cs[n][m];
}
int main() {
    n=read();
    r1=read(); scanf("%s",s1+1);
    r2=read(); scanf("%s",s2+1);
    r3=read(); scanf("%s",s3+1);
    for(int i=1;i<=n;++i) {
        int F=(s1[i]=='1')+(s2[i]=='1')*2+(s3[i]=='1')*4;
        switch(F) {
            case 0:++A;break;
            case 1:++B;break;
            case 2:++C;break;
            case 3:++D;break;
            case 4:++D;break;
            case 5:++C;break;
            case 6:++B;break;
            case 7:++A;break;
        }
    }
    Cs[0][0]=Sm[0][0]=1;
    for(int i=1;i<=n;++i) {
        Cs[i][0]=Cs[i][i]=1;
        for(int j=1;j<i;++j) Cs[i][j]=(Cs[i-1][j-1]+Cs[i-1][j])%mod;
    }
    S(0,-D)=0;
    for(int i=0;i<=A;++i)
        for(int j=0;j<=D;++j)
            S(0,-D)=(S(0,-D)+1ll*Cs[A][i]*Cs[D][j])%mod;
    for(int a=1;a<=A+D;++a) {
        S(a,-D)=S(a-1,-D);
        for(int i=0;i<=A;++i)
            if(a-i>0)
                S(a,-D)=(S(a,-D)-1ll*calc(A,i)*calc(D,a-i-1)%mod+mod)%mod;
    }
    for(int d=-D;d<A;++d) {
        for(int i=A+D;i>=0;--i)
            psum[i]=(psum[i+1]+1ll*calc(A,i)*calc(D,i-d))%mod;
        for(int a=0,lim;a<=A+D;++a) {
            lim=(a+d)/2;
            if((a+d)%2==1) ++lim;
            S(a,d+1)=(S(a,d)-psum[mx(lim,0)]+mod)%mod;
        }
    }
    int ans=0,pnow=1,lim1,lim2;
    for(int i=0;i<=B;++i)
        for(int j=0;j<=C;++j) {
            lim1=r3-(i+j);lim2=mx(r1-(B-i+j),r2-(i+C-j))-D;
            ans=(ans+1ll*Cs[B][i]*Cs[C][j]%mod*S(mx(lim1+1,0),mx(lim2+1,-D)))%mod;
        }
    for(int i=1;i<=n;++i) pnow=2ll*pnow%mod;
    ans=(pnow-ans+mod)%mod;
    printf("%d\n",ans);
    return 0;
}