#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
#include <bitset>

using namespace std;
using lint = long long int;
long long int INF = 1001001001001001LL;
int inf = 1000000007;
long long int MOD = 1000000007LL;
double PI = 3.1415926535897932;

template<typename T1,typename T2>inline void chmin(T1 &a,const T2 &b){if(a>b) a=b;}
template<typename T1,typename T2>inline void chmax(T1 &a,const T2 &b){if(a<b) a=b;}

#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()

/* do your best */

vector<lint> culc(vector<lint> a) {
  int n = a.size();
  lint m = 100 * 1000;
  vector<vector<lint>> dp(n + 1, vector<lint> (m + 1, 0));
  dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= m; j++) {
      if (dp[i][j] == 0) continue;

      dp[i + 1][j] += dp[i][j];
      dp[i + 1][j] %= MOD;

      dp[i + 1][j + a[i]] += dp[i][j];
      dp[i + 1][j + a[i]] %= MOD;
    }
  }
  return dp[n];
}

// 抽象累積和
// 構築 O(n), get O(1)

template<typename T>
struct CumSum{
private: 
  size_t n;
  vector<T> dat;

public: 
  CumSum(const vector<T> &v){
    n = v.size();
    dat.resize(n + 1, 0);
    for(size_t i = 0; i < n; i++){
      dat[i + 1] = dat[i] + v[i];
      dat[i + 1] %= MOD;
    }
  }

  T get(size_t r) const { // 0-indexed, [0. r)
    return dat[r];
  }

  T get(size_t l, size_t r){ // 0-indexed, [l, r)
    return (dat[r] - dat[l] + MOD) % MOD;
  }
};

int main() {
  
  int n, m; cin >> n >> m;
  vector<lint> v(n);
  vector<lint> r(m);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> r[i];
  }

  lint a, b; cin >> a >> b;

  auto dp1 = culc(v);
  auto dp2 = culc(r);

  for (int i = 1; i <= 100 * 1000; i++) {
    dp1[i] += dp1[i - 1];
    dp1[i] %= MOD;
  }

  // 抵抗を決め打ち


  lint ans = 0;
  for (int r = 1; r <= 100 * 1000; r++) {
    // 範囲が決ま
    lint lb = a * r;
    lint rb = b * r;
    if (100 * 1000 < lb) {
      continue;
    }
    rb = min(rb, 100 * 1000LL);

    lint sum = (dp1[rb] - dp1[lb - 1] + MOD) % MOD;
    ans += sum * dp2[r];
    ans %= MOD;

  }

  cout << ans << endl;
  return 0;
}