/*

感觉做这题只要对矩阵乘法理解的稍微一点就能做出来

对于每一行构造一个矩阵

A = a 1

    0 b

列与列之间的矩阵为

B = c 1

    0 d

最终答案为

$A^{n - 1}B A^{n - 1}B$

把$A^{n-1}B$看成一项进行快速幂即可

maya把数据范围看漏了1e6个0。。。。。。。

好像把快速幂换成十进制快速幂就行了

*/

#include<bits/stdc++.h>

#define LL long long

#define int long long

const int MAXN = 1e6 + , mod = 1e9 + , L = ;

using namespace std;

char t1[MAXN], t2[MAXN];

int N[MAXN], M[MAXN], a, b, c, d, n, m;

int Mod(int x, int y) {

    if(1ll * x * y > mod) return 1ll * x * y % mod;

    else return 1ll * x * y;

}

int add(int x, int y) {

    if(x + y > mod) return x + y - mod;

    else return x + y;

}

struct Matrix {

    int m[][];

    Matrix() {

        memset(m, , sizeof(m));

    }

    Matrix operator * (const Matrix &rhs) const {

        Matrix ans;

        /*for(int k = 1; k <= L; k++)

            for(int i = 1; i <= L; i++)

                for(int j = 1; j <= L; j++)

                    (ans.m[i][j] +=  1ll * m[i][k] * rhs.m[k][j] % mod) %= mod;*/

        ans.m[][] = add(Mod(m[][], rhs.m[][]), Mod(m[][], rhs.m[][]));

        ans.m[][] = add(Mod(m[][], rhs.m[][]), Mod(m[][], rhs.m[][]));

        ans.m[][] = add(Mod(m[][], rhs.m[][]), Mod(m[][], rhs.m[][]));

        ans.m[][] = add(Mod(m[][], rhs.m[][]), Mod(m[][], rhs.m[][]));

        return ans;

    }

};

Matrix fp(Matrix a, int *p, int len) {

    Matrix base;

    for(int i = ; i <= ; i++) base.m[i][i] = ;

    for(int i = len; i >= ; i--) {

        for(int j = ; j <= p[i]; j++) base = base * a;

        Matrix tmp = a;

        a = a * a; a = a * a; a = a * a; a = a * tmp * tmp;

    }

    return base;

}

int trans(char *s, int l, int *to) {

    for(int i = ; i <= l; i++) to[i] = s[i] - '';

    to[l]--;

    for(int i = l; i >= ; i--)

        if(to[i] < ) to[i - ] += to[i], to[i] =  + to[i];

        else break;

    return l;

}

main() {

//    freopen("a.in", "r", stdin);

    scanf("%s%s%d%d%d%d", t1 + , t2 + , &a, &b, &c, &d);

    n = strlen(t1 + ); m = strlen(t2 + );

    n = trans(t1, n, N); m = trans(t2, m, M);

    Matrix x, y;

    x.m[][] = a; x.m[][] = b;

    x.m[][] = ; x.m[][] = ;

    y.m[][] = c; y.m[][] = d;

    y.m[][] = ; y.m[][] = ;

    Matrix debug = fp(x, M, m) ;

    debug = debug * y;

    Matrix ans = fp(debug, N, n);

    ans = ans * fp(x, M, m);

    cout << (ans.m[][] + ans.m[][]) % mod;

}