1 /*
2 题意：求a(n)=a(n-1)+a(n-2)+a(n-3)+......a(n-k)+... k<20 n<10^18
3 题解：矩阵快速幂，把递推公式写成矩阵的形式，就可以利用快速幂计算
4 时间：2018.07.31
5 */
6
7 #include <bits/stdc++.h>
8 using namespace std;
9
10 typedef long long LL;
11 const int MAXN = 100005;
12 const LL MOD7 = 1e9+7;
13 const LL MOD9 = 1e9+9;
14
15
16 int n,K;
17 LL m;
18
19 struct Matrix
20 {
21 LL a[25][25];
22 int n;
23 Matrix()
24 {
25 memset(a,0LL,sizeof(a));
26 n= 20;
27 }
28 Matrix(int n):n(n)
29 {
30 memset(a,0LL,sizeof(a));
31 }
32 Matrix(const Matrix& T)
33 {
34 this->n = T.n;
35 for (int i=0;i<n;++i)
36 {
37 for (int j=0;j<n;++j)
38 {
39 a[i][j]=T.a[i][j];
40 }
41 }
42 }
43
44 Matrix operator*(const Matrix& T)
45 {
46 Matrix c(n);
47 for (int i=0;i<n;++i)
48 {
49 for (int j=0;j<n;++j)
50 {
51 for (int k=0;k<n;++k)
52 {
53 c.a[i][j]=(c.a[i][j] + a[i][k]*T.a[k][j] % MOD9) % MOD9;
54 }
55 }
56 }
57 return c;
58 }
59
60 Matrix operator+(const Matrix &T)
61 {
62 Matrix c(n);
63 for (int i=0;i<n;++i)
64 {
65 for (int j=0;j<n;++j)
66 {
67 c.a[i][j] = (c.a[i][j]+a[i][j]+T.a[i][j])%MOD9;
68 }
69 }
70 return c;
71 }
72
73 Matrix operator=(const Matrix &T)
74 {
75 this->n=T.n;
76 for (int i=0;i<n;++i)
77 {
78 for (int j=0;j<n;++j)
79 {
80 a[i][j]=T.a[i][j];
81 }
82 }
83 }
84
85 void ones()
86 {
87 for (int i=0;i<n;++i) a[i][i]=1;
88 }
89
90 void Print()
91 {
92 for (int i=0;i<n;++i)
93 {
94 for (int j=0;j<n;++j)
95 {
96 printf("%lld ",a[i][j]);
97 }
98 printf("\n");
99 }
100 }
101 };
102
103 Matrix Pow(Matrix a, LL b)
104 {
105 Matrix res(a.n);
106 res.ones();
107 Matrix ans(a);
108 while (b)
109 {
110 if (b&1) res=res*ans;
111 ans=ans*ans;
112 b>>=1;
113 }
114 return res;
115 }
116
117 int b[25];
118
119 int main()
120 {
121 #ifndef ONLINE_JUDGE
122 freopen("test.txt","r",stdin);
123 #endif // ONLINE_JUDGE
124 while (scanf("%d%lld%d",&n,&m,&K)!=-1)
125 {
126 for (int i=1;i<=n;++i) scanf("%d",&b[n-i+1]);
127 Matrix a(n);
128 for (int i=0;i<n-1;++i) a.a[i][i+1]=1;
129 int ck;
130 for (int i=1;i<=K;++i)
131 {
132 scanf("%d",&ck);
133 a.a[ck-1][0]+=1;
134 }
135 // a.Print();
136 if (m<=n) {
137 printf("%d\n",b[n-m+1]);
138 continue;
139 }
140 a=Pow(a,m-n);
141 // a.Print();
142 LL ans=0;
143 for (int i=0;i<n;++i)
144 ans = (ans + b[i+1]*a.a[i][0]) % MOD9;
145 printf("%lld\n",ans);
146 }
147 return 0;
148 }
