#include using namespace std; typedef long long ll; #define F first #define S second #define pii pair #define eb emplace_back #define all(v) v.begin(), v.end() #define rep(i, n) for (int i = 0; i < (n); ++i) #define rep3(i, l, n) for (int i = l; i < (n); ++i) #define sz(v) (int)v.size() #define endl '\n' const int inf = 1000000007; const ll INF = 1e18; // int mod = 998244353; int mod = 1000000007; #define abs(x) (x >= 0 ? x : -(x)) #define lb(v, x) (int)(lower_bound(all(v), x) - v.begin()) #define ub(v, x) (int)(upper_bound(all(v), x) - v.begin()) template inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; } template inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; } ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); } ll lcm(ll a, ll b) { return a / gcd(a, b) * b; } ll pow(ll a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; } ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; } template ostream& operator << (ostream& os, const pair& p) { os << p.F << " " << p.S; return os; } template ostream& operator << (ostream& os, const vector& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; } template inline istream& operator >> (istream& is, vector& v) { rep(j, sz(v)) is >> v[j]; return is; } template inline void add(T &a, T2 b) { a += b; if (a >= mod) a -= mod; } void solve(); int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); cout << fixed << setprecision(10); int T; T = 1; while (T--) solve(); } struct mint { ll x; mint (ll x = 0) : x(x >= 0 ? x % mod : (mod + x % mod) % mod) {} mint operator -() const { return mint(-x); } mint& operator += (const mint a) { // コンストラクタの、modで割った余りがくる if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator -= (const mint a) { if ((x += -a.x + mod) >= mod) x -= mod; return *this; } mint& operator *= (const mint a) { (x *= a.x) %= mod; return *this; } mint operator + (const mint a) const { mint res(*this); return res += a; // ? } mint operator - (const mint a) const { mint res(*this); return res -= a; } mint operator * (const mint a) const { mint res(*this); return res *= a; } mint pow(ll t) const { // mint(a).pow(t) tは余りとったりしたらあかん if (t == 0) return 1; mint a = pow(t >> 1); // 再帰 a *= a; if (t & 1) a *= *this; return a; } // 逆元 mint inv() const { return pow(mod - 2); } mint& operator /= (const mint a) { return (*this) *= a.inv(); } mint operator / (const mint a) const { mint res(*this); return res /= a; } friend ostream &operator << (ostream &s, mint a) { return s << a.x; } }; // https://yukicoder.me/submissions/474362 const ll N = 100005; mint dp[101][N], dp2[101][N]; void solve() { ll n, m; cin >> n >> m; vector v(n); cin >> v; vector r(m); cin >> r; ll a, b; cin >> a >> b; dp[0][0] = 1; rep(i, n) rep(j, N) { dp[i + 1][j] += dp[i][j]; if (j + v[i] < N) dp[i + 1][j + v[i]] += dp[i][j]; } dp[n][0] = 0; rep3(j, 1, N) dp[n][j] += dp[n][j - 1]; dp2[0][0] = 1; rep(i, m) rep(j, N) { dp2[i + 1][j] += dp2[i][j]; if (j + r[i] < N) dp2[i + 1][j + r[i]] += dp2[i][j]; } dp2[m][0] = 0; rep3(j, 1, N) dp2[m][j] += dp2[m][j - 1]; mint ans = 0; rep3(V, 1, N) { // V/B ≤ R ll R2 = (V + b - 1) / b; ll R = V / a; ans += (dp[n][V] - dp[n][V - 1]) * (dp2[m][R] - dp2[m][R2 - 1]); } cout << ans << endl; }