#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <stack>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <cstdlib>
#include <bitset>
#include <tuple>
#include <assert.h>
#include <deque>
#include <bitset>
#include <iomanip>
#include <limits>
#include <chrono>
#include <random>
#include <array>
#include <unordered_map>
#include <functional>
#include <complex>
#include <numeric>

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

constexpr long long MAX = 5100000;
constexpr long long INF = 1LL << 60;
constexpr int inf = 1 << 28;
constexpr long long mod = 1000000007LL;
//constexpr long long mod = 998244353LL;

using namespace std;
typedef unsigned long long ull;
typedef long long ll;

struct mint {
	long long x;
	mint(long long x = 0) :x((x% mod + mod) % mod) {}
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= 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 {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	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;
	}
};

constexpr int mx = 100000;

mint dp1[101][100001];
mint dp2[101][100001];

int main()
{
	/*
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	*/

	int n, m; scanf("%d %d", &n, &m);
	vector<int> v(n); for (int i = 0; i < n; i++) scanf("%d", &v[i]);
	vector<int> r(m); for (int i = 0; i < m; i++) scanf("%d", &r[i]);
	ll A, B; scanf("%lld %lld", &A, &B);
	dp1[0][0] = dp2[0][0] = 1;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j <= mx; j++) {
			if (j + v[i] <= mx) dp1[i + 1][j + v[i]] += dp1[i][j];
			dp1[i + 1][j] += dp1[i][j];
		}
	}
	for (int i = 0; i < m; i++) {
		for (int j = 0; j <= mx; j++) {
			if (j + r[i] <= mx) dp2[i + 1][j + r[i]] += dp2[i][j];
			dp2[i + 1][j] += dp2[i][j];
		}
	}
	vector<mint> sum(mx + 2); 
	for (int i = 0; i <= mx; i++) sum[i + 1] = sum[i] + dp1[n][i];
	mint res = 0;
	for (ll R = 1; R <= mx; R++) {
		ll left = R * A;
		if (left > mx) break;
		ll right = min(R * B, (ll)mx);
		//cout << left << " " << right << endl;
		//if((dp2[m][R] * (sum[right] - sum[left - 1])).x > 0) cout<<R<<" "<<left<<" "<<right<< " " << (dp2[m][R] * (sum[right] - sum[left - 1])).x<<endl;
		res += dp2[m][R] * (sum[right + 1] - sum[left]);
	}
	cout << res.x << "\n";
	return 0;
}