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.
import java.io.IOException; 
import java.io.InputStream; 
import java.io.PrintWriter;
import java.util.*;

public class d{ 

	static class Node{
		int id, bit, k;
		int ans = 0;
		Node(int id, int bit, int k){this.id=id;this.bit=bit;this.k=k;}
	}
	static class Bitset{
		int bit, sum;
		Bitset(int bit, int sum){this.bit=bit;this.sum=sum;}
	}

	static void solve(){ 
		int n = ni(), m=ni(), q=ni();
		int[] w = nia(n);
		List<Bitset> bitsets = new ArrayList<>();
		for(int i=0;i<(1<<n);++i){
			int res=0;
			for(int j=0;j<n;++j){
				if(((i>>j)&1)==1)res+=w[n-1-j];
			}
			bitsets.add(new Bitset(i, res));
		}

		Collections.sort(bitsets, (a,b)->a.sum-b.sum);
		int[] S = new int[(1<<n)];
		for(int i=0;i<m;++i){
			String s = ns();
			int res = 0;
			for(int j=0;j<n;++j)res=(res*2)+s.charAt(j)-'0';
			S[res]++;
		}
		List<Node> nodes = new ArrayList<>();
		for(int i=0;i<q;++i){
			int res = 0;
			String s =ns();
			int k = ni();
			for(int j=0;j<n;++j)res=(res*2)+s.charAt(j)-'0';
			nodes.add(new Node(i, res, k));
		}
		Collections.sort(nodes, (a,b)->a.k-b.k);
		int[] cnt = new int[1<<n];
		int bindex=0;
		for(int i=0;i<q;++i){
			Node node = nodes.get(i);
			while(bindex<(1<<n) &&bitsets.get(bindex).sum<=node.k){
				Bitset bitset = bitsets.get(bindex);
				for(int j=0;j<(1<<n);++j){
					cnt[bitset.bit^j^((1<<n)-1)]+=S[j];
				}
				++bindex;
			}
			node.ans = cnt[node.bit];
		}
		Collections.sort(nodes, (a,b)->a.id-b.id);
		for(int i=0;i<q;++i)out.println(nodes.get(i).ans);

	} 
 
 
 
 
	public static void main(String[] args){ 
		 solve(); 
		 out.flush(); 
	 } 
	 private static InputStream in = System.in; 
	 private static PrintWriter out = new PrintWriter(System.out); 
 
	 private static final byte[] buffer = new byte[1<<15]; 
	 private static int ptr = 0; 
	 private static int buflen = 0; 
	 private static boolean hasNextByte(){ 
		 if(ptr<buflen)return true; 
		 ptr = 0; 
		 try{ 
			 buflen = in.read(buffer); 
		 } catch (IOException e){ 
			 e.printStackTrace(); 
		 } 
		 return buflen>0; 
	 } 
	 private static int readByte(){ if(hasNextByte()) return buffer[ptr++]; else return -1;} 
	 private static boolean isSpaceChar(int c){ return !(33<=c && c<=126);} 
	 private static int skip(){int res; while((res=readByte())!=-1 && isSpaceChar(res)); return res;} 
 
	 private static double nd(){ return Double.parseDouble(ns()); } 
	 private static char nc(){ return (char)skip(); } 
	 private static String ns(){ 
		 StringBuilder sb = new StringBuilder(); 
		 for(int b=skip();!isSpaceChar(b);b=readByte())sb.append((char)b); 
		 return sb.toString(); 
	 } 
	 private static int[] nia(int n){ 
		 int[] res = new int[n]; 
		 for(int i=0;i<n;++i)res[i]=ni(); 
		 return res; 
	 } 
	 private static long[] nla(int n){ 
		 long[] res = new long[n]; 
		 for(int i=0;i<n;++i)res[i]=nl(); 
		 return res; 
	 } 
	 private static int ni(){ 
		 int res=0,b; 
		 boolean minus=false; 
		 while((b=readByte())!=-1 && !((b>='0'&&b<='9') || b=='-')); 
		 if(b=='-'){ 
			 minus=true; 
			 b=readByte(); 
		 } 
		 for(;'0'<=b&&b<='9';b=readByte())res=res*10+(b-'0'); 
		 return minus ? -res:res; 
	 } 
	 private static long nl(){ 
		 long res=0,b; 
		 boolean minus=false; 
		 while((b=readByte())!=-1 && !((b>='0'&&b<='9') || b=='-')); 
		 if(b=='-'){ 
			 minus=true; 
			 b=readByte(); 
		 } 
		 for(;'0'<=b&&b<='9';b=readByte())res=res*10+(b-'0'); 
		 return minus ? -res:res; 
	} 
} 

