use std::io::*;
const MOD: usize = 1_000_000_007;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let m: usize = itr.next().unwrap().parse().unwrap();

    let v: Vec<usize> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();
    let r: Vec<usize> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();
    let a: usize = itr.next().unwrap().parse().unwrap();
    let b: usize = itr.next().unwrap().parse().unwrap();

    // 小問題１: 電圧の和がxになるものの組み合わせの数
    let mut dp: Vec<usize> = vec![0; 110000];
    dp[0] = 1;
    for i in 0..n {
        for j in (0..101010).rev() {
            dp[j + v[i]] = (dp[j + v[i]] + dp[j]) % MOD;
        }
    }
    let mut dp2: Vec<usize> = vec![0; 110000];
    dp2[0] = 1;
    for i in 0..m {
        for j in (0..101010).rev() {
            dp2[j + r[i]] = (dp2[j + r[i]] + dp2[j]) % MOD;
        }
    }
    for i in 0..101010 {
        dp[i + 1] += dp[i];
        dp[i + 1] %= MOD;
    }

    // a * sum(r) <= sum(v) <= b * sum(r)
    let mut ans = 0;
    for sr in 0..101010 {
        let min = a * sr;
        let max = b * sr;
        if max > 101010 {
            continue;
        }
        if min == 0 {
            ans = (ans + dp2[sr] * dp[max] % MOD) % MOD;
        } else {
            ans = (ans + dp2[sr] * (dp[max] + MOD - dp[min - 1]) % MOD) % MOD;
        }
    }
    println!("{}", ans - 1);
}