int n;
LL a[MAXN], r[MAXN];
 
void Extend_Gcd(LL a, LL b, LL &g, LL &x, LL &y)
{
    if (!b) g = a, x = 1, y = 0;
    else { Extend_Gcd(b, a%b, g, y, x); y -= x * (a/b); }
}
 
int num[10];
 
void Extend_Chinese_Remainder(LL &a1, LL &r1)   //计算K mod ai = ri, ri不互素情况
{
    LL a2, r2, x, y, g;
    for (int i = 1; i < 4; i++)
    {
        a2 = a[i]; r2 = r[i];
        Extend_Gcd(a1, a2, g, x, y);
        LL C = r2 - r1, tmp = a2 / g;
        x = C / g * x;
        x = (x%tmp + tmp) % tmp;
        r1 = a1*x + r1;
        a1 = a1 / g * a2;
        r1 = (r1%a1 + a1) % a1;
    }
}
 
vector<int> ans;
 
int main()
{
    //ROP;
    int T;
    scanf("%d", &T);
    while (T--)
    {
        ans.clear();
        scanf("%d", &n);
        for (int i = 0; i < 4; i++) scanf("%lld", &a[i]);
        for (int i = 0; i < n; i++)
        {
            int tmp;
            scanf("%d", &tmp);
            for (int j = 0; j < 4; j++)
                r[3-j] = tmp % 100, tmp /= 100;
            LL a1 = a[0], r1 = r[0];
            Extend_Chinese_Remainder(a1, r1);
            for (int i = 0; i < 3; i++)
                r[2-i] = r1 % 100, r1 /= 100;
            for (int i = 0; i < 3; i++) ans.PB(r[i]);
        }
        int j;
        for (j = SZ(ans)-1; j >= 0; j--)
            if (ans[j] != 27) break;
        for (int i = 0; i <= j; i++)
            if (ans[i] == 27) printf(" ");
            else printf("%c", ans[i] + 'A' - 1);
        puts("");
    }
    return 0;
}