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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Collection;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.stream.Stream;

public class Solution implements Runnable {
	
	static final double time = 1e9;
	static final int MOD = (int) 1e9 + 7;
	static final long mh = Long.MAX_VALUE;
	static final Reader in = new Reader();
	static final PrintWriter out = new PrintWriter(System.out);
	StringBuilder answer = new StringBuilder();

	public static void main(String[] args) {
		new Thread(null, new Solution(), "persefone", 1 << 28).start();
	}

	@Override
	public void run() {
		long start = System.nanoTime();
		solve();		
		printf();
		long elapsed = System.nanoTime() - start;
//		out.println("stamp : " + elapsed / time);
//		out.println("stamp : " + TimeUnit.SECONDS.convert(elapsed, TimeUnit.NANOSECONDS));
		close();
	}
	
	void solve() {
		int n = in.nextInt();
		int m = in.nextInt();
		int q = in.nextInt();
		int[] w = in.nextIntArray(n);
		
		int[] cs = new int[1 << n];
		for (int i = 0; i < m; i++) {
			int j = Integer.parseInt(in.next(), 2);
			cs[j]++;
		}
		
		int totalSum = Arrays.stream(w).sum();
		int[] val = new int[1 << n];
		for (int i = 0; i < 1 << n; i++) {
			int v = i;
			for (int j = 0; j < n; j++) {
				if ((v & 1) == 1) 
					val[i] += w[n - j - 1];
				v >>= 1;
			}
		}
		
		
		int[][] ans = new int[1 << n][];
		int x = 0;
		for (int t = 0; t < q; t++) {
			int i = Integer.parseInt(in.next(), 2);
			int k = in.nextInt();
			if (ans[i] == null) {
				ans[i] = new int[101];
				for (int j = 0; j < 1 << n; j++) {
					if (cs[j] > 0 && (x = totalSum - val[i ^ j]) <= 100) {
						ans[i][x] += cs[j];
					}
				}
				for (x = 1; x <= 100; x++) ans[i][x] += ans[i][x - 1];
			}
			add(ans[i][k], '\n');
		}		
	}
		
	int pow(int a, int b, int p) {
		long ans = 1, base = a;
		while (b > 0){
			if ((b & 1) > 0) {
				ans *= base %= p;
			}
			base *= base %= p;
			b >>= 1;
		}
		return (int)ans;
	}
	
	public static int[] enumLowestPrimeFactors(int n) {
		int tot = 0;
		int[] lpf = new int[n + 1];
		int u = n + 32;
		double lu = Math.log(u);
		int[] primes = new int[(int) (u / lu + u / lu / lu * 1.5)];
		for (int i = 2; i <= n; i++)
			lpf[i] = i;
		for (int p = 2; p <= n; p++) {
			if (lpf[p] == p)
				primes[tot++] = p;
			int tmp;
			for (int i = 0; i < tot && primes[i] <= lpf[p] && (tmp = primes[i] * p) <= n; i++) {
				lpf[tmp] = primes[i];
			}
		}
		return lpf;
	}

	
	public static int[] enumNumDivisorsFast(int n, int[] lpf)
	{
		int[] nd = new int[n+1];
		nd[1] = 1;
		for(int i = 2;i <= n;i++){
			int j = i, e = 0;
			while(j % lpf[i] == 0){
				j /= lpf[i];
				e++;
			}
			nd[i] = nd[j] * (e+1);
		}
		return nd;
	}
	
	boolean isPrime(long n) {
		
		for (int i = 2; 1l * i * i <= n; i++) 
			if (n % i == 0)
				return false;
		return true;
	}
	
	void printf() {
		out.print(answer);
	}
	
	void close() {
		out.close();
	}

	void printf(Stream<?> str) {
		str.forEach(o -> add(o, " "));
		add("\n");
	}

	 
	void printf(Object... obj) {
		printf(false, obj);
	}

	void printfWithDescription(Object... obj) {
		printf(true, obj);
	}
	
		
	private void printf(boolean b, Object... obj) {

		if (obj.length > 1) {
			for (int i = 0; i < obj.length; i++) {
				if (b) add(obj[i].getClass().getSimpleName(), " - ");
				if (obj[i] instanceof Collection<?>) {
					printf((Collection<?>) obj[i]);
				} else if (obj[i] instanceof int[][]) {
					printf((int[][])obj[i]);
				} else if (obj[i] instanceof long[][]) {
					printf((long[][])obj[i]);
				} else if (obj[i] instanceof double[][]) {
					printf((double[][])obj[i]);
				} else printf(obj[i]);
			}
			return;
		}

		if (b) add(obj[0].getClass().getSimpleName(), " - ");
		printf(obj[0]);
	}

	void printf(Object o) {
		if (o instanceof int[])
			printf(Arrays.stream((int[]) o).boxed());
		else if (o instanceof char[])
			printf(new String((char[]) o));
		else if (o instanceof long[])
			printf(Arrays.stream((long[]) o).boxed());
		else if (o instanceof double[])
			printf(Arrays.stream((double[]) o).boxed());
		else if (o instanceof boolean[]) {
			for (boolean b : (boolean[]) o) add(b, " ");
			add("\n");
		}
		else 
			add(o, "\n");
	}

	void printf(int[]... obj) {
		for (int i = 0; i < obj.length; i++) printf(obj[i]);
	}
	
	void printf(long[]... obj) {
		for (int i = 0; i < obj.length; i++) printf(obj[i]);
	}
	
	void printf(double[]... obj) {
		for (int i = 0; i < obj.length; i++) printf(obj[i]);
	}
	
	void printf(boolean[]... obj) {
		for (int i = 0; i < obj.length; i++) printf(obj[i]);
	}
	
	void printf(Collection<?> col) {
		printf(col.stream());
	}

	<T, K> void add(T t, K k) {
		if (t instanceof Collection<?>) {
			((Collection<?>) t).forEach(i -> add(i, " "));
		} else if (t instanceof Object[]) {
			Arrays.stream((Object[]) t).forEach(i -> add(i, " "));
		} else
			add(t);
		add(k);
	}
	
	
	
	<T> void add(T t) {
		answer.append(t);
	}

	@SuppressWarnings("unchecked")
	<T extends Comparable<? super T>> T min(T... t) {
		if (t.length == 0)
			return null;
		T m = t[0];
		for (int i = 1; i < t.length; i++)
			if (t[i].compareTo(m) < 0)
				m = t[i];
		return m;
	}

	@SuppressWarnings("unchecked")
	<T extends Comparable<? super T>> T max(T... t) {
		if (t.length == 0)
			return null;
		T m = t[0];
		for (int i = 1; i < t.length; i++)
			if (t[i].compareTo(m) > 0)
				m = t[i];
		return m;
	}

	int gcd(int a, int b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	long gcd(long a, long b) {
		return (b == 0) ? a : gcd(b, a % b);
	}
	
	long lcm(int a, int b) {
		return (1l * a * b / gcd(a, b));
	}
	
	long lcm(long a, long b) {
		if (a > Long.MAX_VALUE / b) throw new RuntimeException();
		return a * b / gcd(a, b);
	}
	
	// c = gcd(a, b) -> extends gcd method: ax + by = c <----> (b % a)p + q = c;
	int[] ext_gcd(int a, int b) {
		if (b == 0) return new int[] {a, 1, 0 };
		int[] vals = ext_gcd(b, a % b);
		int d = vals[0]; // gcd
	    int p = vals[2]; // 
		int q = vals[1] - (a / b) * vals[2];
		return new int[] { d, p, q };	  
	}
	
	// find any solution of the equation: ax + by = c using extends gcd
	boolean find_any_solution(int a, int b, int c, int[] root) {
		int[] vals = ext_gcd(Math.abs(a), Math.abs(b));
		if (c % vals[0] != 0) return false;
		printf(vals);
		root[0] = c * vals[1] / vals[0];
		root[1] = c * vals[2] / vals[0];
		if (a < 0) root[0] *= -1;
		if (b < 0) root[1] *= -1;
		return true;
	}
	
	int mod(int x) { return x % MOD; }
	
	int mod(int x, int y) { return mod(mod(x) + mod(y)); }
	
	long mod(long x) { return x % MOD; }
	
	long mod (long x, long y) { return mod(mod(x) + mod(y)); }
	
	int lw(long[] f, int l, int r, long k) {
		int R = r, m = 0;
		while (l <= r) {
			m = l + r >> 1;
		    if (m == R) return m;
		    if (f[m] >= k) r = m - 1; else l = m + 1;
		}
		return l;
	}
	
	int up(long[] f, int l, int r, long k) {
		int R = r, m = 0;
		while (l <= r) {
			m = l + r >> 1;
		    if (m == R) return m;
		    if (f[m] > k) r = m - 1; else l = m + 1;
		}
		return l;
	}
	
	int lw(int[] f, int l, int r, int k) {
		int R = r, m = 0;
		while (l <= r) {
			m = l + r >> 1;
		    if (m == R) return m;
		    if (f[m] >= k) r = m - 1; else l = m + 1;
		}
		return l;
	}
	
	int up(int[] f, int l, int r, int k) {
		int R = r, m = 0;
		while (l <= r) {
			m = l + r >> 1;
		    if (m == R) return m;
		    if (f[m] > k) r = m - 1; else l = m + 1;
		}
		return l;
	}

	<K extends Comparable<K>> int lw(List<K> li, int l, int r, K k) {
		int R = r, m = 0;
		while (l <= r) {
			m = l + r >> 1;
		    if (m == R) return m;
		    if (li.get(m).compareTo(k) >= 0) 
		    	r = m - 1;
		    else 
		    	l = m + 1;
		}
		return l;
	}
	
	<K extends Comparable<K>> int up(List<K> li, int l, int r, K k) {
		int R = r, m = 0;
		while (l <= r) {
			m = l + r >> 1;
		    if (m == R) return m;
		    if (li.get(m).compareTo(k) > 0) 
		    	r = m - 1;
		    else 
		    	l = m + 1;
		}
		return l;
	}
		
	<K extends Comparable<K>> int bs(List<K> li, int l, int r, K k) {
		while (l <= r) {
			int m = l + r >> 1;
		    if (li.get(m) == k) return m;
		    else if (li.get(m).compareTo(k) < 0)
		    	l = m + 1;
		    else 
		    	r = m - 1;
		}
		return -1;
	}
	
	long calc(int base, int exponent, int p) {
		if (exponent == 0) return 1;
		if (exponent == 1) return base % p;
		long m = calc(base, exponent >> 1, p);
		if (exponent % 2 == 0) return (m * m) % p;
		return 1l * base * m  % p * m % p;
	}
	
	long calc(int base, long exponent, int p) {
		if (exponent == 0) return 1;
		if (exponent == 1) return base % p;
		long m = calc(base, exponent >> 1, p);
		if (exponent % 2 == 0) return (m * m) % p;
		return 1l * base * m  % p * m % p;
	}
	
	long calc(long base, long exponent, int p) {
		if (exponent == 0) return 1;
		if (exponent == 1) return base % p;
		long m = calc(base, exponent >> 1, p);
		if (exponent % 2 == 0) return (m * m) % p;
		return base * m  % p * m % p;
	}
	
	long power(int base, int exponent) {
		if (exponent == 0) return 1;
		long m = power(base, exponent / 2);
		if (exponent % 2 == 0) return m * m;
		return base * m * m;
	}
	
	void swap(int[] a, int i, int j) {
		a[i] ^= a[j];
		a[j] ^= a[i];
		a[i] ^= a[j];
	}
	
	void swap(long[] a, int i, int j) {
		long tmp = a[i];
		a[i] = a[j];
		a[j] = tmp;
	}
	
	static class Pair<K extends Comparable<? super K>, V extends Comparable<? super V>>
			implements Comparable<Pair<K, V>> {
		private K k;
		private V v;

		Pair() {}

		Pair(K k, V v) {
			this.k = k;
			this.v = v;
		}

		K getK() { return k; }

		V getV() { return v; }

		void setK(K k) { this.k = k; }

		void setV(V v) { this.v = v; }

		void setKV(K k, V v) {
			this.k = k;
			this.v = v;
		}

		@SuppressWarnings("unchecked")
		@Override
		public boolean equals(Object o) {
			if (this == o) return true;
			if (o == null || !(o instanceof Pair)) return false;
			Pair<K, V> p = (Pair<K, V>) o;
			return k.compareTo(p.k) == 0 && v.compareTo(p.v) == 0;
		}

		@Override
		public int hashCode() {
			int hash = 31;
			hash = hash * 89 + k.hashCode();
			hash = hash * 89 + v.hashCode();
			return hash;
		}

		@Override
		public int compareTo(Pair<K, V> pair) {
			return k.compareTo(pair.k) == 0 ? v.compareTo(pair.v) : k.compareTo(pair.k);
		}

		@Override
		public Pair<K, V> clone() {
			return new Pair<K, V>(this.k, this.v);
		}

		@Override
		public String toString() {
			return String.valueOf(k).concat(" ").concat(String.valueOf(v)).concat("\n");
		}
	}

	static class Reader {
		private BufferedReader br;
		private StringTokenizer st;

		Reader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			try {
				while (st == null || !st.hasMoreTokens()) {
					st = new StringTokenizer(br.readLine());
				}
			} catch (IOException e) {
				e.printStackTrace();
			}
			return st.nextToken();
		}

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

		int[] nextIntArray(int n) {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++)
				arr[i] = nextInt();
			return arr;
		}

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

		String nextLine() {
			String s = "";
			try {
				s = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return s;
		}
	}
}