Codeforces 327E Axis Walking【状压Dp+Lowbit优化】

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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, ..., a_n. The sum of those numbers is d. 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 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 b_j more steps on Ox 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 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, ..., a_n (1 ≤ a_i ≤ 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

Output

Input

Output

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.

题目大意：

现在给你N个数，让你将其按照任意顺序排列，需要保证前缀和不能出现规定的K个数中的任意一个。

问有多少种排列方式。

思路：

看到数据范围，N最大是24.

又是统计方案数问题，我们肯定要考虑状压dp.

设定dp【i】表示已经排列完成了状态为i的序列的可行方案数的个数。

那么状态转移方程我们可以O（2^n*n）去暴力转移。

dp【i】+=dp【q】（q=i-(1<<j)）；

虽然CF跑的飞快，但是出题人还是卡了暴力转移。

我们知道lowbit是取一个数最后位的数。

比如lowbit（10）=2；

lowbit（5）=1.

本身我们枚举的就是状态i中包含的每个位子上的1.

那么我们这里可以lowbit优化一下。

这样就能快很多。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<map>
#include<math.h>
using namespace std;
#define ll __int64
const ll mod=1e9+7;
ll a[25];
ll ban[25];
ll dp[(1<<24)+10];
ll num[(1<<24)+10];
ll lowbit(ll num)
{
    return num&(-num);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)scanf("%I64d",&a[i]),num[1<<i]=a[i];
        int end=(1<<n);
        int k;
        scanf("%d",&k);
        memset(ban,-1,sizeof(ban));
        for(int i=0;i<k;i++)
        {
            scanf("%I64d",&ban[i]);
        }
        dp[0]=1;
        for(int i=0;i<end;i++)
        {
            num[i]=num[i-lowbit(i)]+num[lowbit(i)];
            if(num[i]==ban[0]||num[i]==ban[1])continue;
            for(int j=i;j>0;j-=lowbit(j))
            {
                int q=i-lowbit(j);
                dp[i]+=dp[q];
                if(dp[i]>mod)dp[i]-=mod;
            }
        }
        printf("%I64d\n",dp[end-1]);
    }
}