Problem: Iahub wants to meet his girlfriend Iahubina. They both live in Ox axis (the horizontal axis). Iahub lives at point 0 and Iahubina at point d.

Iahub has n positive integers a1, a2, ..., an. The sum of those numbers is d. Suppose p1, p2, ..., pn is a permutation of {1, 2, ..., n}. Then, let b1 = ap1, b2 = ap2 and so on. The array b is called a "route". There are n! different routes, one for each permutation p.

Iahub's travel schedule is: he walks b1 steps on Ox axis, then he makes a break in point b1. Then, he walks b2 more steps on Ox axis and makes a break in point b1 + b2. Similarly, at j-th (1 ≤ j ≤ n) time he walks bj more steps on Ox axis and makes a break in point b1 + b2 + ... + bj.

Iahub is very superstitious and has k integers which give him bad luck. He calls a route "good" if he never makes a break in a point corresponding to one of those k numbers. For his own curiosity, answer how many good routes he can make, modulo 1000000007 (109 + 7). 

Input

The first line contains an integer n (1 ≤ n ≤ 24). The following line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 109). 

The third line contains integer k (0 ≤ k ≤ 2). The fourth line contains k positive integers, representing the numbers that give Iahub bad luck. Each of these numbers does not exceed 109.

Output

Output a single integer — the answer of Iahub's dilemma modulo 1000000007 (109 + 7). 

Examples

Input

3
2 3 5
2
5 7


Output

1


Input

3
2 2 2
2
1 3


Output

6

Note

In the first case consider six possible orderings:

  * [2, 3, 5]. Iahub will stop at position 2, 5 and 10. Among them, 5 is bad luck for him. 
  * [2, 5, 3]. Iahub will stop at position 2, 7 and 10. Among them, 7 is bad luck for him. 
  * [3, 2, 5]. He will stop at the unlucky 5. 
  * [3, 5, 2]. This is a valid ordering. 
  * [5, 2, 3]. He got unlucky twice (5 and 7). 
  * [5, 3, 2]. Iahub would reject, as it sends him to position 5. 



In the second case, note that it is possible that two different ways have the identical set of stopping. In fact, all six possible ways have the same stops: [2, 4, 6], so there's no bad luck for Iahub.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class TaskE implements Runnable {

	private static final int BASE = 12;
	private static final int MASKA = (1<<BASE)-1;
	private static final int MASKA2 = MASKA<<12;
	private static final int MOD = 1000000007;
	
	private int[] ways;
	private int[] sum;
	
	private void solve() {
		int[] pow = new int[n];
		for (int i = 0; i < n; ++i) {
			pow[i] = 1<<i;
		}
		ways = new int[1<<n];
		ways[0] = 1;
		sum = new int[1<<n];
		sum[0] = 0;
		for (int i = 0; i < n; ++i) {
			sum[1<<i] = a[i];
		}
		int[][] c = new int[1<<BASE][];
		int[][] c2 = new int[1<<BASE][];
		for (int i = 0; i < (1<<BASE); ++i) {
			int cnt = 0;
			for (int bit = 0; bit < BASE; ++bit) {
				if (((1<<bit) & i) > 0) {
					++cnt;
				}
			}
			c[i] = new int[cnt];
			c2[i] = new int[cnt];
			cnt = 0;
			for (int bit = 0; bit < BASE; ++bit) {
				if (((1<<bit) & i) > 0) {
					c2[i][cnt] = (i<<12)^(1<<bit<<12);
					c[i][cnt++] = i^(1<<bit);
				}
			}
		}
		for (int i = 1; i < (1<<n); ++i) {
			int prev = i&(i-1);
			sum[i] = sum[prev] + sum[prev^i];
			if (sum[i] > MOD) {
				sum[i] = MOD;
			}
			if (sum[i] != b[0] && sum[i] != b[1]) {
				long cnt = 0;
				int p = i&MASKA;
				int p2 = i&MASKA2;
				for (int x: c[p]) {
					cnt += ways[x|p2];
				}
				p2 >>= 12;
				for (int x: c2[p2]) {
					cnt += ways[p|x];
				}
				ways[i] = (int)(cnt%MOD);
			}
		}
		out.println(ways[(1<<n)-1]);
	}
	
	private void readData() {
		n = in.readInt();
		a = new int[n];
		for (int i = 0; i < n; ++i) {
			a[i] = in.readInt();
		}
		k = in.readInt();
		b = new int[] {-1, -1};
		for (int i = 0; i < k; ++i) {
			b[i] = in.readInt();
		}
	}
	
	private int n;
	private int[] a;
	
	private int k;
	private int[] b;
	
	private void begin() {
		in = new InputReader(System.in);
		out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	}
	
	private void end() {
		out.flush();
	}
	
	private InputReader in;
	private PrintWriter out;
	
	@Override
	public void run() {
		begin();
		readData();
		solve();		
		end();
	}

	public static void main(String[] args) {
		new TaskE().run();
	}
	
	private class InputReader {
		
	    public BufferedReader reader;
	    public StringTokenizer tokenizer;

	    public InputReader(InputStream stream) {
	        reader = new BufferedReader(new InputStreamReader(stream));
	        tokenizer = null;
	    }

	    public String next() {
	        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } 
	            catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }
	    
	    public int readInt() {
	    	return Integer.parseInt(next());
	    }
	    
	    public long readLong() {
	    	return Long.parseLong(next());
	    }
	    
	    public String readString() {
	    	try {
				return reader.readLine();
			} 
	    	catch (IOException e) {				
				e.printStackTrace();
				throw new RuntimeException(e);
			}
	    }
	}

}