#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#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<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> 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<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }
template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T, class T2> 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<int> v(n);
  cin >> v;
  vector<int> 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];
    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];
    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) {
    ll R2 = V / a - 2;
    ll tmp = R2;
    rep3(i, -3, 4) {
      if (tmp + i <= 0) continue;
      if (V / (tmp + i) < a) chmin(R2, tmp + i);
      else chmax(R2, tmp + i);
    }
    ll R = V / b + 2;
    tmp = R;
    rep3(i, -3, 4) {
      if (tmp + i <= 0) continue;
      if ((V + tmp + i - 1) / (tmp + i) > b) chmax(R, tmp + i);
      else chmin(R, tmp + i);
    }
    // cout << V << " " << R2 << " " << R << endl;
    // V/抵抗 の切り捨てでない !! 小数部も考える
    ans += (dp[n][V] - dp[n][V - 1]) * (dp2[m][max(R2, R)] - dp2[m][min(R2, R) - 1]);
  }
  /*
  rep3(R, 1, N) {
    ll l = R * a;
    if (l >= N) break;
    ll r = min(R * b, N - 1);
    ans += (dp2[m][R] - dp2[m][R - 1]) * (dp[n][r] -  dp[n][l - 1]);
  }
  */
  cout << ans << endl;
}