注意到，M 是質數 乘法取個log變加法 也就是取指標

於是對於1 ~M−1 中的每一個數都可以表示成原根的某次冪。

於是乘法可以轉化為原根的冪的加法，

轉移的時候就相當於做多項式乘法了

然後快速冪

又是道數論好題

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;

inline char nc(){
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(ll &x){
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int M=30005;
const int P=1004535809;
const int G=3;

inline ll Pow(ll a,ll b,ll p){
	ll ret=1;
	for (;b;(a*=a)%=p,b>>=1)
		if (b&1)
			(ret*=a)%=p;
	return ret;
}

inline ll Inv(ll x){
	return Pow(x,P-2,P);
}

ll f[100005];

inline ll GetRoot(ll p,ll phi)
{
	int c=0;
	for (int i=2;i*i<=phi;i++)
		if (phi%i==0)
			f[++c]=i,f[++c]=phi/i;
	for (int g=2;g<p;g++)
	{
		int j;
		for (j=1;j<=c;j++)
			if (Pow(g,f[j],p)==1) 
				break;
		if (j==c+1) return g;
	}
	return 0;
}

ll L,w[2][M],R[M];
ll g,ind[M];
ll n,m,X,S,sp; 

inline void NTT(ll *a,int r){  
    for (int i=0;i<n;i++) if (i<R[i]) swap(a[i],a[R[i]]);  
    for (int i=1;i<n;i<<=1)  
        for (int j=0;j<n;j+=(i<<1))  
            for (int k=0;k<i;k++)  
            {  
                ll x=a[j+k],y=w[r][n/(i<<1)*k]*a[j+k+i]%P;  
                a[j+k]=(x+y)%P; a[j+k+i]=(x-y+P)%P;  
            } 
    if (r==1) for (int i=0,inv=Inv(n);i<n;i++) a[i]=a[i]*inv%P;  
}

struct Poly{
	ll a[M];
	Poly() { memset(a,0,sizeof(a)); }
	Poly(ll x) { a[0]=x; }
	ll& operator [] (const int x) { return a[x]; }
	Poly& operator *=(const Poly &t){
		static ll b[M];
        memcpy(b,t.a,sizeof(b));
        NTT(a,0),NTT(b,0);
        for (int i=0;i<n;i++) a[i]=a[i]*b[i]%P;
        NTT(a,1);
        for (int i=m-1;i<=(m-2)<<1;i++)
            (a[i-(m-1)]+=a[i])%=P,a[i]=0;
        return *this;
	}
	friend Poly Pow(Poly a,ll b){
		Poly ret(1);
		for (;b;b>>=1,a*=a)
			if (b&1)
				ret*=a;
		return ret;
	}
}a;

int main()
{
	ll t;
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(sp); read(m); read(X); read(S);
	g=GetRoot(m,m-1);
	for (n=1;n<=m<<1;n<<=1) L++;
	for(int i=0;i<n;i++) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));  
	for (int i=0,now=1;i<m-1;i++,(now*=g)%=m)
		ind[now]=i;
	w[0][0]=w[0][n]=1;
	t=Pow(G,(P-1)/n,P);
	for (int i=1;i<n;i++) w[0][i]=w[0][i-1]*t%P;
	for (int i=0;i<=n;i++) w[1][i]=w[0][n-i];
	for (int i=1;i<=S;i++)
	{
		read(t); if (t) a[ind[t]]=1;
	}
	printf("%lld\n",Pow(a,sp)[ind[X]]);
	return 0;
}