Problem: Iahub wants to meet his girlfriend Iahubina. They both live in Ox axis (the horizontal axis). Iahub lives at point 0 and Iahubina at point d.

Iahub has n positive integers a1, a2, ..., an. The sum of those numbers is d. Suppose p1, p2, ..., pn is a permutation of {1, 2, ..., n}. Then, let b1 = ap1, b2 = ap2 and so on. The array b is called a "route". There are n! different routes, one for each permutation p.

Iahub's travel schedule is: he walks b1 steps on Ox axis, then he makes a break in point b1. Then, he walks b2 more steps on Ox axis and makes a break in point b1 + b2. Similarly, at j-th (1 ≤ j ≤ n) time he walks bj more steps on Ox axis and makes a break in point b1 + b2 + ... + bj.

Iahub is very superstitious and has k integers which give him bad luck. He calls a route "good" if he never makes a break in a point corresponding to one of those k numbers. For his own curiosity, answer how many good routes he can make, modulo 1000000007 (109 + 7). 

Input

The first line contains an integer n (1 ≤ n ≤ 24). The following line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 109). 

The third line contains integer k (0 ≤ k ≤ 2). The fourth line contains k positive integers, representing the numbers that give Iahub bad luck. Each of these numbers does not exceed 109.

Output

Output a single integer — the answer of Iahub's dilemma modulo 1000000007 (109 + 7). 

Examples

Input

3
2 3 5
2
5 7


Output

1


Input

3
2 2 2
2
1 3


Output

6

Note

In the first case consider six possible orderings:

  * [2, 3, 5]. Iahub will stop at position 2, 5 and 10. Among them, 5 is bad luck for him. 
  * [2, 5, 3]. Iahub will stop at position 2, 7 and 10. Among them, 7 is bad luck for him. 
  * [3, 2, 5]. He will stop at the unlucky 5. 
  * [3, 5, 2]. This is a valid ordering. 
  * [5, 2, 3]. He got unlucky twice (5 and 7). 
  * [5, 3, 2]. Iahub would reject, as it sends him to position 5. 



In the second case, note that it is possible that two different ways have the identical set of stopping. In fact, all six possible ways have the same stops: [2, 4, 6], so there's no bad luck for Iahub.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
void imult(long long& lhs, const long long& rhs) {
  lhs = ((lhs % mod) * (rhs % mod)) % mod;
}
void iadd(long long& lhs, const long long& rhs) {
  lhs = ((lhs % mod) + (rhs % mod)) % mod;
}
void isubt(long long& lhs, const long long& rhs) {
  long long ret = (lhs - rhs) % mod;
  lhs = ret + (ret < 0);
}
long long bpw(const long long& lhs, const long long& rhs) {
  if (rhs == 0) return 1;
  long long ret = bpw(lhs, rhs >> 1);
  imult(ret, ret);
  if (rhs & 1) imult(ret, lhs);
  return ret;
}
long long inv(const long long& x) { return bpw(x, mod - 2); }
long long mult(const long long& lhs, const long long& rhs) {
  return ((lhs % mod) * (rhs % mod)) % mod;
}
long long add(const long long& lhs, const long long& rhs) {
  return ((lhs % mod) + (rhs % mod)) % mod;
}
long long subt(const long long& lhs, const long long& rhs) {
  long long ret = (lhs - rhs) % mod;
  return ret + (ret < 0 ? mod : 0);
}
long long div(const long long& lhs, const long long& rhs) {
  return mult(lhs, inv(rhs));
}
const int sz = 26;
long long f[sz], a[sz], n, k, kk[sz];
void prec() {
  f[0] = 1;
  for (int i = 1; i < sz; i++) f[i] = mult(f[i - 1], i);
}
long long calc(const long long& x, long long tsm, int amt, int i) {
  if (i == n) {
    if (tsm == x)
      return mult(f[amt], f[n - amt]);
    else
      return 0;
  } else
    return calc(x, tsm, amt, i + 1) + calc(x, tsm + a[i], amt + 1, i + 1);
}
map<pair<long long, long long>, map<pair<int, int>, int>> mem[2];
int mysz[2], offset[2];
void calcmap(long long sma, long long smb, int amta, int amtb, const int& ind,
             int i) {
  if (i == mysz[ind]) {
    mem[ind][{sma, smb}][{amta, amtb}]++;
  } else {
    calcmap(sma, smb, amta, amtb, ind, i + 1);
    calcmap(sma + a[offset[ind] + i], smb, amta + 1, amtb, ind, i + 1);
    calcmap(sma, smb + a[offset[ind] + i], amta, amtb + 1, ind, i + 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  cin >> k;
  for (int i = 0; i < k; i++) cin >> kk[i];
  prec();
  if (k == 0) {
    cout << f[n] << '\n';
  } else if (k == 1 || kk[0] == kk[1]) {
    cout << subt(f[n], calc(kk[0], 0, 0, 0)) << '\n';
  } else {
    if (kk[0] > kk[1]) swap(kk[0], kk[1]);
    mysz[0] = n / 2;
    mysz[1] = n - mysz[0];
    offset[1] = mysz[0];
    long long tmp = add(calc(kk[0], 0, 0, 0), calc(kk[1], 0, 0, 0));
    for (int i = 0; i < 2; i++) calcmap(0, 0, 0, 0, i, 0);
    long long sect = 0;
    for (auto x : mem[0]) {
      auto mp1 = x.second;
      pair<long long, long long> p = x.first;
      if (mem[1].find({kk[0] - p.first, kk[1] - kk[0] - p.second}) !=
          mem[1].end()) {
        auto mp2 = mem[1][{kk[0] - p.first, kk[1] - kk[0] - p.second}];
        for (auto it1 : mp1) {
          for (auto it2 : mp2) {
            long long mytmp = mult(mult(f[it1.first.first + it2.first.first],
                                        f[it1.first.second + it2.first.second]),
                                   f[n - it1.first.first - it2.first.first -
                                     it1.first.second - it2.first.second]);
            iadd(sect, mult(mytmp, mult(it1.second, it2.second)));
          }
        }
      }
    }
    long long uni = subt(tmp, sect);
    cout << subt(f[n], uni) << '\n';
  }
  return 0;
}