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 JAVA):

Solution:
import java.io.*;
import java.util.*;

import javax.swing.plaf.synth.SynthSpinnerUI;

public class A {

	static int[] cnt, wildCards, pow_3;
	static int w;
	static char[] s;

	static int getWildCardBit(int bit, char gate) {
		if (gate == 'A')
			return bit == 0 ? 2 : 0;
		else if (gate == 'a')
			return bit == 0 ? -1 : 1;
		else if (gate == 'O')
			return bit == 0 ? 0 : -1;
		else if (gate == 'o')
			return bit == 0 ? 1 : 2;
		else if (gate == 'X')
			return bit == 0 ? 0 : 1;
		else
			return bit == 0 ? 1 : 0;

	}

	static long solve(int msk, int wildCard, int idx) {

		if (idx == w)
			return cnt[msk] * 1L * wildCards[wildCard];
		long ans = 0;
		for (int bit = 0; bit <= 1; bit++) {
			int wildCardBit = getWildCardBit(bit, s[idx]);
			if (wildCardBit != -1)
				ans = ans + solve(msk | (bit << idx), wildCard + pow_3[idx] * wildCardBit, idx + 1);

		}
		return ans;
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		w = sc.nextInt();
		int n = sc.nextInt(), m = sc.nextInt();
		cnt = new int[1 << w];
		while (n-- > 0)
			cnt[sc.nextInt()]++;

		int all = (int) Math.pow(3, w);

		pow_3 = new int[w];
		pow_3[0] = 1;
		for (int i = 1; i < w; i++)
			pow_3[i] = 3 * pow_3[i - 1];
		wildCards = new int[all];
		for (int msk = 0; msk < all; msk++) {
			int first = -1;
			int twos = 0;
			int x = msk;
			int binary = 0;
			for (int i = 0; i < w; i++) {
				int bit = x % 3;
				x /= 3;
				if (bit == 2) {
					twos++;
					if (first == -1)
						first = i;
				} else
					binary |= bit << i;

			}

			if (twos == 0) {
				wildCards[msk] = cnt[binary];
				continue;
			}
			int value = pow_3[first];
			if (twos == 1) {
				wildCards[msk] = cnt[binary] + cnt[binary + (1 << first)];
			} else {
				wildCards[msk] = wildCards[msk - 2 * value] + wildCards[msk - value];
			}

		}
		s = new char[w];
		while (m-- > 0) {
			char[] t = sc.nextLine().toCharArray();
			for (int i = 0; i < w; i++)
				s[i] = t[w - 1 - i];
			out.println(solve(0, 0, 0));
		}
		out.close();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(new File((s))));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
	}
}