Problem: Childan is making up a legendary story and trying to sell his forgery — a necklace with a strong sense of "Wu" to the Kasouras. But Mr. Kasoura is challenging the truth of Childan's story. So he is going to ask a few questions about Childan's so-called "personal treasure" necklace.

This "personal treasure" is a multiset S of m "01-strings".

A "01-string" is a string that contains n characters "0" and "1". For example, if n=4, strings "0110", "0000", and "1110" are "01-strings", but "00110" (there are 5 characters, not 4) and "zero" (unallowed characters) are not.

Note that the multiset S can contain equal elements.

Frequently, Mr. Kasoura will provide a "01-string" t and ask Childan how many strings s are in the multiset S such that the "Wu" value of the pair (s, t) is not greater than k. 

Mrs. Kasoura and Mr. Kasoura think that if s_i = t_i (1≤ i≤ n) then the "Wu" value of the character pair equals to w_i, otherwise 0. The "Wu" value of the "01-string" pair is the sum of the "Wu" values of every character pair. Note that the length of every "01-string" is equal to n.

For example, if w=[4, 5, 3, 6], "Wu" of ("1001", "1100") is 7 because these strings have equal characters only on the first and third positions, so w_1+w_3=4+3=7.

You need to help Childan to answer Mr. Kasoura's queries. That is to find the number of strings in the multiset S such that the "Wu" value of the pair is not greater than k.

Input

The first line contains three integers n, m, and q (1≤ n≤ 12, 1≤ q, m≤ 5⋅ 10^5) — the length of the "01-strings", the size of the multiset S, and the number of queries.

The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_i ≤ 100) — the value of the i-th caracter.

Each of the next m lines contains the "01-string" s of length n — the string in the multiset S.

Each of the next q lines contains the "01-string" t of length n and integer k (0≤ k≤ 100) — the query.

Output

For each query, print the answer for this query.

Examples

Input

2 4 5
40 20
01
01
10
11
00 20
00 40
11 20
11 40
11 60


Output

2
4
2
3
4


Input

1 2 4
100
0
1
0 0
0 100
1 0
1 100


Output

1
2
1
2

Note

In the first example, we can get:

"Wu" of ("01", "00") is 40.

"Wu" of ("10", "00") is 20.

"Wu" of ("11", "00") is 0.

"Wu" of ("01", "11") is 20.

"Wu" of ("10", "11") is 40.

"Wu" of ("11", "11") is 60.

In the first query, pairs ("11", "00") and ("10", "00") satisfy the condition since their "Wu" is not greater than 20.

In the second query, all strings satisfy the condition.

In the third query, pairs ("01", "11") and ("01", "11") satisfy the condition. Note that since there are two "01" strings in the multiset, the answer is 2, not 1.

In the fourth query, since k was increased, pair ("10", "11") satisfies the condition too.

In the fifth query, since k was increased, pair ("11", "11") satisfies the condition too.

Solution (in JAVA):

Solution:
public class D {

	public D () {
		int N = sc.nextInt(), M = sc.nextInt(), Q = sc.nextInt();
		int [] W = sc.nextInts();

		int [] S = new int [M];
		for (int i : rep(M))
			S[i] = Integer.parseInt(sc.next(), 2);

		int L = 1 << N;
		int [] Y = new int [L];
		for (int m : rep(L))
			for (int i : rep(N))
				if (has(m, i))
					Y[m] += W[N-1-i];

		int [] C = new int [L];
		for (int m : S)
			++C[m];

		int [][] Z = new int [L][105];
		for (int m : rep(L)) {
			for (int n : rep(L)) {
				int v = m ^ (L - 1 - n);
				if (Y[v] < 105)
					Z[m][Y[v]] += C[n];
			}
			for (int j : rep(1, 105))
				Z[m][j] += Z[m][j-1];
		}

		for (@SuppressWarnings("unused") int q : rep(Q)) {
			int T = Integer.parseInt(sc.next(), 2);
			int K = sc.nextInt();

			int res = Z[T][K];
			print(res);
		}
	}

	///////////////////////////////////////////////////////////////////////////
	private static long and (long x, int i) { return (x & bit(i)); }
	private static long bit (int i) { return (1L << i); }
	private static boolean has(long x, int i) { return and(x, i) != 0; }
	private static int [] rep(int N) { return rep(0, N); }
	private static int [] rep(int S, int T) { if (T <= S) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	////////////////////////////////////////////////////////////////////////////////////
	private final static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static void print (Object o, Object ... A) { IOUtils.print(o, A); }
	private static class IOUtils {
		public static class MyScanner {
			public String next() { newLine(); return line[index++]; }
			public int nextInt() { return Integer.parseInt(next()); }
			public String nextLine() { line = null; return readLine(); }
			public String [] nextStrings() { return split(nextLine()); }
			public int [] nextInts() {
				String [] L = nextStrings();
				int [] res = new int [L.length];
				for (int i = 0; i < L.length; ++i)
					res[i] = Integer.parseInt(L[i]);
				return res;
			}
			//////////////////////////////////////////////
			private boolean eol() { return index == line.length; }
			private String readLine() {
				try {
					return r.readLine();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private final java.io.BufferedReader r;
			private MyScanner () { this(new java.io.BufferedReader(new java.io.InputStreamReader(System.in))); }
			private MyScanner (java.io.BufferedReader r) {
				try {
					this.r = r;
					while (!r.ready())
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine() {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private String [] split(String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build(Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim(String delim, Object o, Object ... A) {
			StringBuilder b = new StringBuilder();
			append(b, o, delim);
			for (Object p : A)
				append(b, p, delim);
			return b.substring(delim.length());
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static void start() { if (t == 0) t = millis(); }
		private static void append(StringBuilder b, Object o, String delim) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					append(b, java.lang.reflect.Array.get(o, i), delim);
			} else if (o instanceof Iterable<?>)
				for (Object p : (Iterable<?>) o)
					append(b, p, delim);
			else {
				if (o instanceof Double)
					o = new java.text.DecimalFormat("#.############").format(o);
				b.append(delim).append(o);
			}
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void print(Object o, Object ... A) { pw.println(build(o, A)); }
		private static void err(Object o, Object ... A) { System.err.println(build(o, A)); }
		private static void exit() {
			IOUtils.pw.close();
			System.out.flush();
			err("------------------");
			err(IOUtils.time());
			System.exit(0);
		}
		private static long t;
		private static long millis() { return System.currentTimeMillis(); }
		private static String time() { return "Time: " + (millis() - t) / 1000.0; }
	}
	public static void main (String[] args) { new D(); IOUtils.exit(); }
}