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:
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.List;

public class Main {

	private static final int MAXN = 5000;
	private static final String NO = "NO";
	private static final String YES = "YES";
	InputStream is;
	PrintWriter out;
	String INPUT = "";

	private static final long MOD = 1000000007L;

	void solve() {
		int n = ni();
		int m = ni();
		int q = ni();
		int a[] = na(n);
		int[] cnt = new int[1 << n];
		for (int i = 0; i < m; i++) {
			cnt[Integer.valueOf(ns(), 2)]++;
		}

//		tr(a, cnt);
		int cost[][] = new int[1 << n][101];
		for (int i = 0; i < (1 << n); i++) {
			for (int j = 0; j < (1 << n); j++) {
				int k = i ^ j;
				int s = 0;
				for (int f = 0; f < n; f++) {
					if ((k & (1 << f)) == 0)
						s += a[n - 1 - f];
				}
//				tr(i, j, s);
				if (s < 101)
					cost[i][s] += cnt[j];
			}
//			tr(i, cost[i]);
//			cost[i][0] = 0;
			for (int j = 1; j <= 100; j++)
				cost[i][j] = cost[i][j - 1] + cost[i][j];
//			tr(i, cost[i]);
		}
//		tr(cost);

		while (q-- > 0) {
			int i = Integer.valueOf(ns(), 2);
			out.println(cost[i][ni()]);
		}
	}

	long power(long a, long b) {
		long x = 1, y = a;
		while (b > 0) {
			if (b % 2 != 0) {
				x = (x * y) % MOD;
			}
			y = (y * y) % MOD;
			b /= 2;
		}
		return x % MOD;
	}

	private long gcd(long a, long b) {
		while (a != 0) {
			long tmp = b % a;
			b = a;
			a = tmp;
		}
		return b;
	}

	void run() throws Exception {
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if (!INPUT.isEmpty())
			tr(System.currentTimeMillis() - s + "ms");
	}

	public static void main(String[] args) throws Exception {
		new Main().run();
	}

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	private boolean vis[];

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}

	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != '
									// ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n) {
			if (!(isSpaceChar(b)))
				buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private List<Integer> na2(int n) {
		List<Integer> a = new ArrayList<Integer>();
		for (int i = 0; i < n; i++)
			a.add(ni());
		return a;
	}

	private int[][] na(int n, int m) {
		int[][] a = new int[n][];
		for (int i = 0; i < n; i++)
			a[i] = na(m);
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long[] nl(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private static void tr(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	public class Pair<K, V> {

		/**
		 * Key of this <code>Pair</code>.
		 */
		private K key;

		/**
		 * Gets the key for this pair.
		 * 
		 * @return key for this pair
		 */
		public K getKey() {
			return key;
		}

		/**
		 * Value of this this <code>Pair</code>.
		 */
		private V value;

		/**
		 * Gets the value for this pair.
		 * 
		 * @return value for this pair
		 */
		public V getValue() {
			return value;
		}

		/**
		 * Creates a new pair
		 * 
		 * @param key   The key for this pair
		 * @param value The value to use for this pair
		 */
		public Pair(K key, V value) {
			this.key = key;
			this.value = value;
		}

		/**
		 * <p>
		 * <code>String</code> representation of this <code>Pair</code>.
		 * </p>
		 *
		 * <p>
		 * The default name/value delimiter '=' is always used.
		 * </p>
		 *
		 * @return <code>String</code> representation of this <code>Pair</code>
		 */
		@Override
		public String toString() {
			return key + "=" + value;
		}

		/**
		 * <p>
		 * Generate a hash code for this <code>Pair</code>.
		 * </p>
		 *
		 * <p>
		 * The hash code is calculated using both the name and the value of the
		 * <code>Pair</code>.
		 * </p>
		 *
		 * @return hash code for this <code>Pair</code>
		 */
		@Override
		public int hashCode() {
			// name's hashCode is multiplied by an arbitrary prime number (13)
			// in order to make sure there is a difference in the hashCode between
			// these two parameters:
			// name: a value: aa
			// name: aa value: a
			return key.hashCode() * 13 + (value == null ? 0 : value.hashCode());
		}

		/**
		 * <p>
		 * Test this <code>Pair</code> for equality with another <code>Object</code>.
		 * </p>
		 *
		 * <p>
		 * If the <code>Object</code> to be tested is not a <code>Pair</code> or is
		 * <code>null</code>, then this method returns <code>false</code>.
		 * </p>
		 *
		 * <p>
		 * Two <code>Pair</code>s are considered equal if and only if both the names and
		 * values are equal.
		 * </p>
		 *
		 * @param o the <code>Object</code> to test for equality with this
		 *          <code>Pair</code>
		 * @return <code>true</code> if the given <code>Object</code> is equal to this
		 *         <code>Pair</code> else <code>false</code>
		 */
		@Override
		public boolean equals(Object o) {
			if (this == o)
				return true;
			if (o instanceof Pair) {
				Pair pair = (Pair) o;
				if (key != null ? !key.equals(pair.key) : pair.key != null)
					return false;
				if (value != null ? !value.equals(pair.value) : pair.value != null)
					return false;
				return true;
			}
			return false;
		}
	}

}