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.util.*;
import java.io.*;
public class helloworld {
 static long fact[];
 static long max;
 static int ans;
	static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 
	
	static long getSum(int index , long BITree[]) 
    { 
		
        long  sum = 0; // Iniialize result 
      
        // index in BITree[] is 1 more than 
        // the index in arr[] 
    
      
        // Traverse ancestors of BITree[index] 
        while(index>0) 
        { 
            // Add current element of BITree  
            // to sum 
            sum += BITree[index]; 
      
            // Move index to parent node in  
            // getSum View 
            index -= index & (-index); 
        } 
        return sum; 
    } 
  
	static int getMid(int s, int e) 
    { 
        return s + (e - s) / 2; 
    } 
  
    /* 
    * A recursive function to get the sum  
    of values in given range of the array. 
    * The following are parameters for this function. 
    *  
    * st -> Pointer to segment tree 
    * node -> Index of current node in 
    *         the segment tree. 
    * ss & se -> Starting and ending indexes 
    *         of the segment represented 
    *         by current node, i.e., st[node] 
    * l & r -> Starting and ending indexes 
    *         of range query 
    */
    static int MaxUtil(int[] st, int ss, int se,  
                        int l, int r, int node)  
    { 
          
        // If segment of this node is completely 
        // part of given range, then return 
        // the max of segment 
        if (l <= ss && r >= se) 
            return st[node]; 
  
        // If segment of this node does not 
        // belong to given range 
        if (se < l || ss > r) 
            return -1; 
  
        // If segment of this node is partially 
        // the part of given range 
        int mid = getMid(ss, se); 
  
        return Math.max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), 
                    MaxUtil(st, mid + 1, se, l, r, 2 * node + 2)); 
    } 
  
    /* 
    * A recursive function to update the  
    nodes which have the given index in their 
    * range. The following are parameters  
    st, ss and se are same as defined above 
    * index -> index of the element to be updated. 
    */
    static void updateValue(int arr[], int[] st, int ss, 
                    int se, int index, int value, int node) 
    { 
        if (index < ss || index > se) 
        { 
            System.out.println("Invalid Input"); 
            return; 
        } 
  
        if (ss == se) 
        { 
  
            // update value in array and in segment tree 
            arr[index] = value; 
            st[node] = value; 
        }  
        else
        { 
            int mid = getMid(ss, se); 
  
            if (index >= ss && index <= mid) 
                updateValue(arr, st, ss, mid, index, value, 2 * node + 1); 
            else
                updateValue(arr, st, mid + 1, se, index, value, 2 * node + 2); 
  
            st[node] = Math.max(st[2 * node + 1], st[2 * node + 2]); 
        } 
        return; 
    } 
  
    // Return max of elements in range from 
    // index l (query start) to r (query end). 
    static int getMax(int[] st, int n, int l, int r) 
    { 
  
        // Check for erroneous input values 
        if (l < 0 || r > n - 1 || l > r)  
        { 
            System.out.printf("Invalid Input\n"); 
            return -1; 
        } 
  
        return MaxUtil(st, 0, n - 1, l, r, 0); 
    } 
  
    // A recursive function that constructs Segment 
    // Tree for array[ss..se]. si is index of 
    // current node in segment tree st 
    static int constructSTUtil(int arr[], int ss, 
                        int se, int[] st, int si)  
    { 
          
        // If there is one element in array, store 
        // it in current node of segment tree and return 
        if (ss == se)  
        { 
            st[si] = arr[ss]; 
            return arr[ss]; 
        } 
  
        // If there are more than one elements, then 
        // recur for left and right subtrees and 
        // store the max of values in this node 
        int mid = getMid(ss, se); 
  
        st[si] = Math.max(constructSTUtil(arr, ss, mid, st, si * 2 + 1), 
                constructSTUtil(arr, mid + 1, se, st, si * 2 + 2)); 
  
        return st[si]; 
    } 
  
    /* 
    * Function to construct segment tree from 
    given array. This function allocates 
    * memory for segment tree. 
    */
    static int[] constructST(int arr[], int n) 
    { 
  
        // Height of segment tree 
        int x = (int) Math.ceil(Math.log(n) / Math.log(2)); 
  
        // Maximum size of segment tree 
        int max_size = 2 * (int) Math.pow(2, x) - 1; 
  
        // Allocate memory 
        int[] st = new int[max_size]; 
  
        // Fill the allocated memory st 
        constructSTUtil(arr, 0, n - 1, st, 0); 
  
        // Return the constructed segment tree 
        return st; 
    } 
    public static void updateBIT(int n, int index,  
                                        long val , long BITree[] ) 
    { 
        // index in BITree[] is 1 more than  
        // the index in arr[] 
       
        // Traverse all ancestors and add 'val' 
        while(index <= n) 
        { 
        // Add 'val' to current node of BIT Tree 
        BITree[index] += val; 
      
        // Update index to that of parent  
        // in update View 
        index += index & (-index); 
        } 
    } 
    static long mod = 1000000007;
   static long pow(long a , long b , long c)
   {
	   
	   if(b==0)
		   return 1;
	   
	   long ans = pow(a,b/2,c);
	   if(b%2 == 0)
		   return ans*ans%c;
	   
	   return ans*ans%c*a%c;
   }
   
   static long modInverse(long a , long b)
   {
	   return pow(a,b-2,mod);
   }
   
   
   static double d(int x1 , int y1 , int x2 , int y2)
   {
	   double d1 = x1-x2;
	   double d2 = y1-y2;
	   
	   return Math.sqrt(d1*d1-d2*d2);
   }
  

static int low[];
static int p[];
static int parent(int x , int k , int dp[][])
{
	if(k==0)
		return x;
while(k > 0)
{
	if(x == -1)
		break;
	
	int up = low[k];
	k -= p[up];
	x = dp[x][up];
}
return x;
}


static void d(int n , LinkedList<Integer> arr[] , boolean visited[] , int dp[][] , int h[] , int p,int lev)
{
	visited[n] = true;
	int lim = low[visited.length];
	
	dp[n][0] = p;
	int temp = n;
	for(int j = 1 ; j <= lim ; j++)
	{
		if(dp[n][j-1] == -1)
			break;
		
		dp[n][j] = dp[dp[n][j-1]][j-1];
	}
	
	h[n] = lev;
	
	for(Integer i : arr[n])
	{
		if(!visited[i])
		{
			d(i,arr,visited,dp,h,n,lev+1);
		}
	}
}

static int lca(int a , int b , int dp[][] , int h[])
{
	if(h[a] > h[b])
	{
		a = parent(a,h[a]-h[b],dp);
		
	}
	
	else if(h[b] > h[a])
	{
		b = parent(b,h[b]-h[a],dp);
	}
	
	if(a==b)
		return a;
	
	
	int l = 1 , r = h.length;
	
	while(l < r)
	{
		int mid = (l+r)/2;
		if(parent(a,mid,dp) == parent(b,mid,dp))
			r = mid;
		
		else
			l = mid+1;
	}
	
	return parent(a,l,dp);
}


static void dfs(int n , LinkedList<Integer> arr[] , boolean visited[] , int dp[] , int idx[] , int ans[])
{
	visited[n] = true;
	for(Integer i : arr[n])
	{
		if(!visited[i])
		{
			dfs(i,arr,visited,dp,idx,ans);
			dp[n] += dp[i];
		}
	}
	if(n != 1)
	ans[idx[n]] = dp[n];
	//System.out.println(dp[n] + " " + idx[n]);
}

public static void main(String []args) throws IOException
{
    Scanner sc = new Scanner(System.in);
	int n = sc.nextInt();
	int m = sc.nextInt();
	int q = sc.nextInt();
	
	int w[] = new int[n];
	for(int i = 0 ; i < n ; i++)
	{
		w[i] = sc.nextInt();
	}
	int arr[] = new int[1<<12];
	for(int i = 0 ; i < 1<<12 ; i++)
	{
		int sum = 0;
		for(int j = 0 ; j < 12 ; j++)
		{
			if((i&(1<<j)) == 0 && j < n)
				sum += w[n-1-j];
		}
		arr[i] = sum;
	}
	sc.nextLine();
	int mult[] = new int[1<<12];
    for(int i = 0 ; i < m ; i++)
    {
    	String s = sc.nextLine();
    	int bit = 0;
    	for(int j = s.length()-1 ; j >= 0 ; j--)
    	{
    		if(s.charAt(j) == '1')
    		{
    			bit += 1 << (s.length()-1-j);
    		}
    	}
    	mult[bit]++;
    }
    
    int ans[][] = new int[1<<12][1201];
    for(int i = 0 ; i < 1<<12 ; i++)
    {
    	for(int j = 0 ; j < 1<<12 ; j++)
    	{
    		
    		
    		ans[i][arr[i^j]] += mult[j];
    		
    	}
    }
    
   // System.out.println(arr[2] + " " + arr[3]);
    for(int i = 0 ; i < 1<<12 ; i++)
    {
    	for(int j = 1 ; j <= 1200 ; j++)
    	{
    		ans[i][j] += ans[i][j-1];
    	}
    }
    StringBuffer str = new StringBuffer("");
    while(q-- > 0)
    {
    	String in = sc.nextLine();
    	String s = in.substring(0,in.indexOf(' '));
    	int k = Integer.parseInt(in.substring(in.indexOf(' ')+1));
    	int bit = 0;
    	for(int i = s.length()-1 ; i >= 0 ; i--)
    	{
    		if(s.charAt(i) == '1')
    			bit += 1<<(s.length()-1-i);
    	}
    	str.append(ans[bit][Math.min(1200, k)]);
    	str.append(System.getProperty("line.separator"));
    }
    System.out.println(str);
}
}