Problem: A pitch-black room

When I woke up, Mr. A was in a pitch-black room. Apparently, Mr. A got lost in a dungeon consisting of N rooms. You couldn't know which room A got lost in, but fortunately he got a map of the dungeon. Let's show A the way to go and lead to a bright room.

It is known that M of the N rooms are pitch black, and the D_1, D_2, ..., and D_Mth rooms are pitch black, respectively. Also, there are exactly K one-way streets from all the rooms in order, and the roads from the i-th room are v_ {i, 1}, v_ {i, 2}, ..., v_ {i, respectively. , K} Connected to the third room. You can follow the a_1, a_2, ..., a_lth roads in order from the room you are in for Mr. A. However, when Mr. A reaches the bright room, he ignores the subsequent instructions. You cannot know the information of the room where Mr. A is currently before and after the instruction, so you must give the instruction sequence so that you can reach the bright room no matter which room you are in. Answer the length of the shortest of such instructions.

Constraints

* 2 ≤ N ≤ 100
* 1 ≤ M ≤ min (16, N − 1)
* 1 ≤ K ≤ N
* 1 ≤ D_i ≤ N
* D_i are all different
* 1 ≤ v_ {i, j} ≤ N
* All dark rooms can reach at least one bright room



Input Format

Input is given from standard input in the following format.


NMK
D_1 D_2 ... D_M
v_ {1,1} v_ {1,2} ... v_ {1,K}
v_ {2,1} v_ {2,2} ... v_ {2, K}
...
v_ {N, 1} v_ {N, 2} ... v_ {N, K}


Output Format

Print the answer in one line.

Sample Input 1


4 2 2
1 2
twenty four
3 1
4 2
13


Sample Output 1


2


<image>
If you give the instruction 1, 1

* If A's initial position is room 1, the second move will reach room 3.
* If A's initial position is room 2, the first move will reach room 3.



Sample Input 2


3 2 2
1 2
twenty three
1 2
twenty one


Sample Output 2


3


<image>
If you give instructions 2, 1, 2

* If A's initial position is room 1, the first move will reach room 3.
* If A's initial position is room 2, the third move will reach room 3.



Sample Input 3


6 3 3
one two Three
4 1 1
2 5 2
3 3 6
4 4 4
5 5 5
6 6 6


Sample Output 3


3


<image>
Beware of unconnected cases and cases with self-edges and multiple edges.





Example

Input

4 2 2
1 2
2 4
3 1
4 2
1 3


Output

2

Solution (in CPP):

Solution:
#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include "bits/stdc++.h"

using namespace std;
using ll = long long int;

#define debug(v) {printf("L%d %s > ",__LINE__,#v);cout<<(v)<<endl;}
#define debugv(v) {printf("L%d %s > ",__LINE__,#v);for(auto e:(v)){cout<<e<<" ";}cout<<endl;}
#define debuga(m,w) {printf("L%d %s > ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}
#define debugaa(m,h,w) {printf("L%d %s >\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}
#define ALL(v) (v).begin(),(v).end()
#define repeat(cnt,l) for(remove_const<decltype(l)>::type cnt=0;(cnt)<(l);++(cnt))
#define rrepeat(cnt,l) for(auto cnt=(l)-1;0<=(cnt);--(cnt))
#define iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))
#define diterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);--(cnt))
const ll MD = 1000000007ll; const long double PI = 3.1415926535897932384626433832795L;
inline void assert_call(bool assertion, function<void()> f) { if (!assertion) { cerr << "assertion fault:" << endl; f(); abort(); } }
template<typename T1, typename T2> inline ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << '(' << p.first << ':' << p.second << ')'; return o; }
template<typename Vec> inline ostream& _ostream_vecprint(ostream &o, const Vec& p) { o << '['; for (auto& e : p) o << e << ','; o << ']'; return o; }
template<typename T> inline ostream& operator<<(ostream& o, const vector<T>& v) { return _ostream_vecprint(o, v); }
template<typename T, size_t S> inline ostream& operator<<(ostream& o, const array<T, S>& v) { return _ostream_vecprint(o, v); }
template<typename T> inline T& maxset(T& to, const T& val) { return to = max(to, val); }
template<typename T> inline T& minset(T& to, const T& val) { return to = min(to, val); }
void bye(string s, int code = 0) { cout << s << endl; exit(code); }
mt19937_64 randdev(8901016);
template<typename T> inline T rand(T l, T h) { return uniform_int_distribution<T>(l, h)(randdev); }
template<> inline double rand<double>(double l, double h) { return uniform_real_distribution<double>(l, h)(randdev); }
template<> inline float rand<float>(float l, float h) { return uniform_real_distribution<float>(l, h)(randdev); }

#if defined(_WIN32) || defined(_WIN64)
#define getchar_unlocked _getchar_nolock
#define putchar_unlocked _putchar_nolock
#elif defined(__GNUC__)
#else
#define getchar_unlocked getchar
#define putchar_unlocked putchar
#endif
namespace {
#define isvisiblechar(c) (0x21<=(c)&&(c)<=0x7E)
    class MaiScanner {
    public:
        template<typename T> void input_integer(T& var) {
            var = 0; T sign = 1;
            int cc = getchar_unlocked();
            for (; cc<'0' || '9'<cc; cc = getchar_unlocked())
                if (cc == '-') sign = -1;
            for (; '0' <= cc && cc <= '9'; cc = getchar_unlocked())
                var = (var << 3) + (var << 1) + cc - '0';
            var = var * sign;
        }
        inline int c() { return getchar_unlocked(); }
        inline MaiScanner& operator>>(int& var) { input_integer<int>(var); return *this; }
        inline MaiScanner& operator>>(long long& var) { input_integer<long long>(var); return *this; }
        inline MaiScanner& operator>>(string& var) {
            int cc = getchar_unlocked();
            for (; !isvisiblechar(cc); cc = getchar_unlocked());
            for (; isvisiblechar(cc); cc = getchar_unlocked())
                var.push_back(cc);
            return *this;
        }
        template<typename IT> void in(IT begin, IT end) { for (auto it = begin; it != end; ++it) *this >> *it; }
    };
    class MaiPrinter {
    public:
        template<typename T>
        void output_integer(T var) {
            if (var == 0) { putchar_unlocked('0'); return; }
            if (var < 0)
                putchar_unlocked('-'),
                var = -var;
            char stack[32]; int stack_p = 0;
            while (var)
                stack[stack_p++] = '0' + (var % 10),
                var /= 10;
            while (stack_p)
                putchar_unlocked(stack[--stack_p]);
        }
        inline MaiPrinter& operator<<(char c) { putchar_unlocked(c); return *this; }
        inline MaiPrinter& operator<<(int var) { output_integer<int>(var); return *this; }
        inline MaiPrinter& operator<<(long long var) { output_integer<long long>(var); return *this; }
        inline MaiPrinter& operator<<(char* str_p) { while (*str_p) putchar_unlocked(*(str_p++)); return *this; }
        inline MaiPrinter& operator<<(const string& str) {
            const char* p = str.c_str();
            const char* l = p + str.size();
            while (p < l) putchar_unlocked(*p++);
            return *this;
        }
        template<typename IT> void join(IT begin, IT end, char sep = '\n') { for (auto it = begin; it != end; ++it) *this << *it << sep; }
    };
}
MaiScanner scanner;
MaiPrinter printer;




int N, M, K;
vector<int> arcs[20];

int cvidx[111], revidx[111];

int dp[1 << 16];

int main() {

    scanner >> N >> M >> K;

    fill(cvidx, cvidx + N, -1); // -1: 明るい部屋

    repeat(i, M) {
        int d; scanner >> d; --d;
        cvidx[d] = 0;
    }
    {
        int idx = 0;
        repeat(i, N) {
            if (cvidx[i] != -1) {
                cvidx[i] = idx;
                revidx[idx] = i;
                ++idx;
            }
        }
    }

    repeat(i, N) {
        // ライトが付いた部屋の繊維は要らない
        if (cvidx[i] == -1) { repeat(j, K) { int v; scanner >> v; } continue; }
        // 変換後のインデックスで辺を張る
        arcs[cvidx[i]].reserve(K);
        repeat(j, K) {
            int v; scanner >> v; --v;
            arcs[cvidx[i]].push_back(cvidx[v]);
        }

    }

    fill(dp, dp + (1 << M), int(1e9));
    dp[(1 << M) - 1] = 0;
    queue<int> qu;
    qu.push((1 << M) - 1);

    while(!qu.empty()) {
        int b = qu.front(); qu.pop();

        repeat(j, K) { // j番目の通路
            int bit = b;
            int to = 0;
            repeat(i, M) {
                if (bit&(1 << i)) {
                    int k = arcs[i][j];
                    if (k >= 0) to |= (1 << k);
                }
            }
            if (dp[to] > dp[b] + 1) {
                dp[to] = dp[b] + 1;
                qu.push(to);
            }
        }
    }

    cout << dp[0] << endl;

    return 0;
}