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

public class E {
	static int MOD = 1000000007;
	public static void main(String[] args) throws Exception{
		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int n = in.nextInt();
		int[] sum = new int[1<<n];
		int mask = 1;
		for(int i = 0; i < n; i++){
			sum[mask] = in.nextInt();
			mask <<= 1;
		}
		
		for(int i = 1; i < (1<<n); i++){
			int lowBit = i & -i;
			int prev = i - lowBit;
			sum[i] = sum[prev] + sum[lowBit];
			//cap so no overflow, because we only care for sum of 2 numbers <= 10^9
			//and we only need sum[i] != any of the bad numbers
			if(sum[i] > 1000000000){
				sum[i] = 1000000001;
			}
		}
		
		int k = in.nextInt();
		int b1 = -1;
		int b2 = -1;
		if(k >= 1) b1 = in.nextInt();
		if(k == 2) b2 = in.nextInt();
		
		//dp[state] = number of good paths that can be formed using the 1's
		//reuse sum
		int[] dp = sum;
		dp[0] = 1;
		
		for(int i = 1; i < (1<<n); i++){
			//sum will be untouched before we alter it for dp[i]
			if(sum[i] == b1 || sum[i] == b2){
				dp[i] = 0;
			}else{
				long ans = 0;
				//try all the possible ending sequence of the good paths
				mask = 1;
				for(int j = 0; j < n; j++){
					int prev = i ^ mask;
					if(prev < i){
						ans += dp[prev];
					}
					mask <<= 1;
				}
				dp[i] = (int)( ans % MOD);
			}
		}
		
		out.println(dp[(1<<n)-1]);
		out.close();
	}

	static class InputReader {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}
		
		
		public int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			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 long nextLong() {
			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 String nextToken() {
			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 boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return isWhitespace(c);
		}

		public static boolean isWhitespace(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}


}