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:
/* Author: Ronak Agarwal, reader part not written by me*/
import java.io.* ; import java.util.* ;
import static java.lang.Math.min ; import static java.lang.Math.max ;
import static java.lang.Math.abs ; import static java.lang.Math.log ;
import static java.lang.Math.pow ; import static java.lang.Math.sqrt ;
/* Thread is created here to increase the stack size of the java code so that recursive dfs can be performed */
public class Codeshefcode{
	public static void main(String[] args) throws IOException{
		new Thread(null,new Runnable(){	public void run(){ exec_Code() ;} },"Solver",1l<<27).start() ;
	}
	static void exec_Code(){
		try{
			Solver Machine = new Solver() ;
			Machine.Solve() ; Machine.Finish() ;}
		catch(Exception e){ e.printStackTrace() ; print_and_exit() ;}
		catch(Error e){ e.printStackTrace() ; print_and_exit() ; }
	}
	static void print_and_exit(){ System.out.flush() ; System.exit(-1) ;}
}
/* Implementation of the Reader class */
class Reader { 
	private InputStream mIs;
	private byte[] buf = new byte[1024];
	private int curChar,numChars;
	public Reader() { this(System.in); } 
	public Reader(InputStream is) { mIs = is;} 
	public int read() {
		if (numChars == -1) throw new InputMismatchException();    
		if (curChar >= numChars) {
			curChar = 0;
			try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}
			if (numChars <= 0) return -1;            
		}
		return buf[curChar++];
	} 
	public String nextLine(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isEndOfLine(c));
		return res.toString() ;
	} 
	public String s(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		}while (!isSpaceChar(c));
		return res.toString();
	} 
	public long l(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		long res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
			res *= 10 ; res += c - '0' ; c = read();
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public int i(){
		int c = read() ;
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		int res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
			res *= 10 ; res += c - '0' ; c = read() ;
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } 
	public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } 
}   
/* All the useful functions,constants,renamings are here*/
class Template{
	/* Constants Section */
	final int INT_MIN = Integer.MIN_VALUE ; final int INT_MAX = Integer.MAX_VALUE ;
	final long LONG_MIN = Long.MIN_VALUE ; final long LONG_MAX = Long.MAX_VALUE ;
	static long MOD = 1000000007 ;
	static Reader ip = new Reader() ;	
	static PrintWriter op = new PrintWriter(System.out) ;
	/* Methods for writing */
	static void p(Object o){ op.print(o) ; }
	static void pln(Object o){ op.println(o) ;}
	static void Finish(){ op.flush(); op.close(); }
	/* Implementation of operations modulo MOD */
	static long inv(long a){ return powM(a,MOD-2) ; }
	static long m(long a,long b){ return (a*b)%MOD ; }
	static long d(long a,long b){ return (a*inv(b))%MOD ; }
	static long powM(long a,long b){
		if(b<0) return powM(inv(a),-b) ;
		long ans=1 ;
		for( ; b!=0 ; a=(a*a)%MOD , b/=2) if((b&1)==1) ans = (ans*a)%MOD ;
		return ans ;
	}
	/* Renaming of some generic utility classes */
	final static class mylist extends ArrayList<pair>{}
	final static class myset extends TreeSet<pair>{}
	final static class mystack extends Stack<Integer>{}
	final static class mymap extends TreeMap<Integer,Integer>{}
	
}
/* Implementation of the pair class, useful for every other problem */
class pair implements Comparable<pair>{
	int x ; long y ;
	pair(int _x,long _y){ x=_x ; y=_y ;} 
	public int compareTo(pair p){
		return (this.x<p.x ? -1 : (this.x>p.x ? 1 : (this.y<p.y ? -1 : (this.y>p.y ? 1 : 0)))) ;
	}
}
/* Main code starts here */
class Solver extends Template{
 	public void Solve() throws IOException{
 		int n,m,q;
 		n = ip.i(); m = ip.i(); q = ip.i();
 		int w[] = new int[n];
 		for(int i=0 ; i<n ; i++) w[i] = ip.i();
 		int N = 1<<n;
 		int cnt[] = new int[N];
 		for(int i=0 ; i<m ; i++){
 			char cr[] = ip.s().toCharArray();
 			int num=0;
 			for(int j=0 ; j<n ; j++) if(cr[j]=='1') num|=(1<<j);
 			cnt[num]++;
 		}
 		int wu[] = new int[N];
 		for(int i=0 ; i<N ; i++){
 			int sum=0;
 			for(int j=0 ; j<n ; j++) if(((1<<j)&i)==0) sum+=w[j];
 			wu[i] = sum;
 		}
 		long res[][] = new long[N][101];
 		for(int i=0 ; i<N ; i++){
 			for(int j=0 ; j<N ; j++){ 
 				int val = wu[i^j];
 				if(val<=100) res[i][val]+=cnt[j];
 			}
 			for(int j=1 ; j<=100 ; j++) res[i][j] += res[i][j-1];
 		}
 		while(q--!=0){
 			char cr[] = ip.s().toCharArray();
 			int num=0;
 			for(int j=0 ; j<n ; j++) if(cr[j]=='1') num|=(1<<j);
 			int k = ip.i();
 			pln(res[num][k]);
 		} 
 	}
}