Problem: Alice has a computer that operates on w-bit integers. The computer has n registers for values. The current content of the registers is given as an array a_1, a_2, …, a_n. 

The computer uses so-called "number gates" to manipulate this data. Each "number gate" takes two registers as inputs and calculates a function of the two values stored in those registers. Note that you can use the same register as both inputs.

Each "number gate" is assembled from bit gates. There are six types of bit gates: AND, OR, XOR, NOT AND, NOT OR, and NOT XOR, denoted "A", "O", "X", "a", "o", "x", respectively. Each bit gate takes two bits as input. Its output given the input bits b_1, b_2 is given below:

\begin{matrix} b_1 & b_2 & A & O & X & a & o & x \\\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\\ 0 & 1 & 0 & 1 & 1 & 1 & 0 & 0 \\\ 1 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\\ \end{matrix} 

To build a "number gate", one takes w bit gates and assembles them into an array. A "number gate" takes two w-bit integers x_1 and x_2 as input. The "number gate" splits the integers into w bits and feeds the i-th bit of each input to the i-th bit gate. After that, it assembles the resulting bits again to form an output word. 

For instance, for 4-bit computer we might have a "number gate" "AXoA" (AND, XOR, NOT OR, AND). For two inputs, 13 = 1101_2 and 10 = 1010_2, this returns 12 = 1100_2, as 1 and 1 is 1, 1 xor 0 is 1, not (0 or 1) is 0, and finally 1 and 0 is 0. 

You are given a description of m "number gates". For each gate, your goal is to report the number of register pairs for which the "number gate" outputs the number 0. In other words, find the number of ordered pairs (i,j) where 1 ≤ i,j ≤ n, such that w_k(a_i, a_j) = 0, where w_k is the function computed by the k-th "number gate".

Input

The first line contains three integers: w, n, and m~(1 ≤ w ≤ 12, 1 ≤ n ≤ 3⋅ 10^4, 1 ≤ m ≤ 5⋅ 10^4) — the word size, the number of variables, and the number of gates.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^w) — the value of variables stored in the registers.

Each of the next m lines contains a string g_j~(|g_j| = w) with a description of a single gate. Each character of g_j is one of "A", "O", "X", "a", "o", "x".

Output

Print m lines. The i-th line should contain the number of ordered pairs of variables for which the i-th gate returns zero.

Examples

Input

4 3 1
13 10 6
AXoA


Output

3


Input

1 7 6
0 1 1 0 1 0 0
A
O
X
a
o
x


Output

40
16
25
9
33
24


Input

6 2 4
47 12
AOXaox
AAaaAA
xxxxxx
XXXXXX


Output

2
3
0
2


Input

2 2 2
2 0
xO
Ox


Output

2
0

Note

In the first test case, the inputs in binary are 1101, 1010, 0110. The pairs that return 0 are (13, 6), (6, 13), and (6, 6). As it was already mentioned in the problem statement, 13 ⊕ 10 = 10 ⊕ 13 = 12. The other pairs are 13 ⊕ 13 = 11, 10 ⊕ 10 = 8 and 10 ⊕ 6 = 6 ⊕ 10 = 4. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,mmx,sse,sse2,tune=native")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T, typename S>
inline ostream& operator<<(ostream& os, const pair<T, S> p) {
  cout << "[" << p.first << ";" << p.second << "]";
  return os;
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
template <typename T>
inline ostream& operator<<(ostream& os, const set<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
template <typename T, typename S>
inline ostream& operator<<(ostream& os, const map<T, S>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
template <typename T>
inline ostream& operator<<(ostream& os, const multiset<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
namespace ___debug {
struct DebugStream {
 private:
  bool is_first;

 public:
  DebugStream(bool _is_first) : is_first(_is_first) {}
  template <typename T>
  DebugStream operator<<(const T& value) const {
    assert(0);
    if (is_first) cout << "[DBG] ";
    cout << value;
    return DebugStream(false);
  };
  template <typename T>
  DebugStream printArray(T* l, T* r) {
    assert(0);
    if (is_first) cout << "[DBG] ";
    while (l != r) {
      cout << (*l);
      ++l;
      if (l == r) {
        cout << '\n';
      } else {
        cout << ' ';
      }
    }
    return DebugStream(false);
  }
};
DebugStream instance(true);
};  // namespace ___debug
const int INT_INF = (int)(2e9);
const ll LL_INF = (ll)(2e18);
const int NIL = -1;
static mt19937 _g(time(nullptr));
inline ll randint(ll a, ll b) {
  ll w = (_g() << 31LL) ^ _g();
  return a + w % (b - a + 1);
}
inline void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
};
template <typename T>
inline T sign(T x) {
  return T(x > 0) - T(x < 0);
}
template <typename T>
inline T fetch() {
  T ret;
  cin >> ret;
  return ret;
}
template <typename T>
inline vector<T> fetch_vec(int sz) {
  vector<T> ret(sz);
  for (auto& elem : ret) cin >> elem;
  return ret;
}
const int MAXN = (int)1e6;
int w, n, m;
int cnt[MAXN], freq[MAXN];
int pw3[MAXN], mod3[MAXN];
void init() {
  pw3[0] = 1;
  for (int i = 1; i < 20; ++i) pw3[i] = 3 * pw3[i - 1];
  mod3[0] = 0;
  for (int i = 1; i < MAXN; ++i)
    mod3[i] = (mod3[i - 1] == 2 ? 0 : mod3[i - 1] + 1);
}
inline int getTrit(int msk, int pos) { return mod3[msk / pw3[pos]]; }
inline int setTrit(int msk, int pos, int val) {
  int cur = getTrit(msk, pos);
  msk += pw3[pos] * (val - cur);
  return msk;
}
inline string fmt(int msk, int sz) {
  string rez(sz, '?');
  for (int i = 0; i < sz; ++i) {
    int v = getTrit(msk, i);
    if (v) rez[i] = '0' + (v - 1);
  }
  return rez;
}
inline int get(char cmd, int a, int b) {
  if (cmd == 'A')
    return a & b;
  else if (cmd == 'O')
    return a | b;
  else if (cmd == 'X')
    return a ^ b;
  else if (cmd == 'a')
    return (a & b) ^ 1;
  else if (cmd == 'o')
    return (a | b) ^ 1;
  else if (cmd == 'x')
    return (a ^ b) ^ 1;
  else
    assert(false);
}
int dp[256][2];
void solve() {
  init();
  cin >> w >> n >> m;
  for (int i = 0; i < n; ++i) {
    int a = fetch<int>();
    int msk = 0;
    for (int i = 0; i < w; ++i) msk = setTrit(msk, i, ((a >> i) & 1) + 1);
    ++cnt[msk];
    ++freq[a];
  }
  int pw = 1;
  for (int j = 0; j < w; ++j) {
    for (int msk = 0; msk < pw3[w]; ++msk) {
      int v = mod3[msk / pw];
      if (v) {
        int msk2 = msk - pw * v;
        cnt[msk2] += cnt[msk];
      }
    }
    pw = pw + (pw << 1);
  }
  for (auto& cmd : string("AOXaox")) {
    int code = (int)cmd;
    for (int bt = 0; bt < 2; ++bt) {
      int zero = 0, one = 0;
      if (get(cmd, bt, 0) == 0) zero = 1;
      if (get(cmd, bt, 1) == 0) one = 1;
      if (zero && one)
        dp[code][bt] = 0;
      else if (zero)
        dp[code][bt] = 1;
      else if (one)
        dp[code][bt] = 2;
      else
        dp[code][bt] = NIL;
    }
  }
  for (int que = 0; que < m; ++que) {
    auto func = fetch<string>();
    reverse(func.begin(), func.end());
    ll answ = 0;
    for (int a = 0; a < (1 << w); ++a) {
      if (!freq[a]) continue;
      int msk = 0, good = 1;
      int pw = 1;
      for (int j = 0; j < w; ++j) {
        int cur = dp[(int)func[j]][(a >> j) & 1];
        if (cur == NIL) {
          good = 0;
          break;
        }
        msk += cur * pw;
        pw = pw + (pw << 1);
      }
      if (good) answ += cnt[msk] * 1LL * freq[a];
    }
    cout << answ << '\n';
  }
}
int main() {
  fast_io();
  solve();
  return 0;
}