codeforce 327E.Axis Walking

Iahub wants to meet his girlfriend Iahubina. They both live inOx axis (the horizontal axis). Iahub lives at point 0 and Iahubina at pointd.

Iahub has n positive integers a₁,a₂, ...,a_n. The sum of those numbers isd. Suppose p₁,p₂, ...,p_n is a permutation of{1, 2, ..., n}. Then, let b₁ = a_p1,b₂ = a_p2 and so on. The array b is called a "route". There are n! different routes, one for each permutationp.

Iahub's travel schedule is: he walks b₁ steps onOx axis, then he makes a break in point b₁. Then, he walksb₂ more steps onOx axis and makes a break in point b₁ + b₂. Similarly, atj-th (1 ≤ j ≤ n) time he walksb_j more steps onOx axis and makes a break in point b₁ + b₂ + ... + b_j.

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 thosek numbers. For his own curiosity, answer how many good routes he can make, modulo1000000007 (10⁹ + 7).

Input

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

The third line contains integer k (0 ≤ k ≤ 2). The fourth line containsk positive integers, representing the numbers that give Iahub bad luck. Each of these numbers does not exceed10⁹.

Output

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

Examples

Input

32 3 525 7

Output

1

Input

32 2 221 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.

#include <cstdio>#include <cstring>#include <algorithm>#define lowbit(x) (x & (-x))//求出2^p(其中p: x 的二进制表示数中， 右向左数第一个1的位置)//如6的二进制表示为110，向左数第零个为0，第一个为1，则p=1，故Lowbit(6) = 2^1 = 2using namespace std;int const MOD = 1e9 + 7;int const MAX = (1 << 24) + 1;long long sum[MAX],dp[MAX];int a[MAX], no[2];int main(){    int n,k;    scanf("%d",&n);    for(int i=0; i<n; i++)    {        scanf("%d",&a[i]);        sum[1<<i]=a[i];    }    scanf("%d",&k);    for(int i=0; i<k; i++)        scanf("%d",&no[i]);    dp[0]=1;    for(int i=1; i<(1<<n); i++)//sum记录的的是选的几个1后的和//    {        sum[i]=sum[i&~lowbit(i)]+sum[lowbit(i)];//选01中1的个数和计算//        if(sum[i]==no[0]||sum[i]==no[1])            continue;        for(int j=i;j!=0;j-=lowbit(j))        {            dp[i]+=dp[i&~lowbit(j)];            if(dp[i]>MOD)                dp[i]-=MOD;        }    }    printf("%I64d\n",dp[(1<<n)-1]);}

这个状态压缩出乎意料，一千六百万数组没有炸内存，遍历一遍中间还有个小循环没有超时，毕竟CF跑的快。

学会了lowbit这个用法，