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:
#include <vector>
#include <iostream>
#include <utility>
#include <algorithm>
#include <string>
#include <deque>
#include <queue>
#include <tuple>
#include <queue>
#include <functional>
#include <cmath>
#include <iomanip>
#include <map>
#include <set>
#include <numeric>
#include <unordered_map>
#include <unordered_set>
#include <complex>
#include <iterator>
#include <array>
#include <memory>
//cin.sync_with_stdio(false);
//streambuf
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
using vi = vector<int>;
using vll = vector<ll>;
using vpii = vector<pii>;
using vpll = vector<pll>;
using ti3 = tuple<int, int, int>;
using vti3 = vector<ti3>;
template<class T, int s>using va = vector<array<T, s>>;
template<class T, class T2> using umap = unordered_map<T, T2>;
template<class T> using uset = unordered_set<T>;
template<class T, class S> void cmin(T &a, const S &b) { if (a > b)a = b; }
template<class T, class S> void cmax(T &a, const S &b) { if (a < b)a = b; }
#define ALL(a) a.begin(),a.end()
#define rep(i,a) for(int i=0;i<a;i++)
#define rep1(i,a) for(int i=1;i<=a;i++)
#define rrep(i,a) for(int i=(a)-1;i>=0;i--)
#define rrep1(i,a) for(int i=a;i;i--)
const ll mod = 1000000007;
#define REP rep
template<class T>using heap = priority_queue<T, vector<T>, greater<T>>;
template<class T>using pque = priority_queue<T, vector<T>, function<T(T, T)>>;
template <class T>
inline void hash_combine(size_t & seed, const T & v) {
	hash<T> hasher;
	seed ^= hasher(v) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
}
namespace std {
	template<typename S, typename T> struct hash<pair<S, T>> {
		inline size_t operator()(const pair<S, T> & v) const {
			size_t seed = 0;
			hash_combine(seed, v.first);
			hash_combine(seed, v.second);
			return seed;
		}
	};
	// Recursive template code derived from Matthieu M.
	template <class Tuple, size_t Index = std::tuple_size<Tuple>::value - 1>
	struct HashValueImpl{
		static void apply(size_t& seed, Tuple const& tuple){
			HashValueImpl<Tuple, Index - 1>::apply(seed, tuple);
			hash_combine(seed, std::get<Index>(tuple));
		}
	};
	template <class Tuple>
	struct HashValueImpl<Tuple, 0>{
		static void apply(size_t& seed, Tuple const& tuple){
			hash_combine(seed, std::get<0>(tuple));
		}
	};
	template <typename ... TT>
	struct hash<std::tuple<TT...>>{
		size_t operator()(std::tuple<TT...> const& tt) const{
			size_t seed = 0;
			HashValueImpl<std::tuple<TT...> >::apply(seed, tt);
			return seed;
		}
	};
}
template<class T>int id(vector<T> &a, T b) {
	return lower_bound(ALL(a), b) - a.begin();
}
ll pow(ll base, ll i, ll mod) {
	ll a = 1;
	while (i) {
		if (i & 1) {
			a *= base;
			a %= mod;
		}
		base *= base;
		base %= mod;
		i /= 2;
	}
	return a;
}
ll gcd(ll a, ll b) {
	while (b) {
		ll c = a%b;
		a = b;
		b = c;
	}
	return a;
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b)*b;
}
int popcnt(unsigned long long a) {
	a = (a & 0x5555555555555555) + (a >> 1 & 0x5555555555555555);
	a = (a & 0x3333333333333333) + (a >> 2 & 0x3333333333333333);
	a = (a & 0x0f0f0f0f0f0f0f0f) + (a >> 4 & 0x0f0f0f0f0f0f0f0f);
	a = (a & 0x00ff00ff00ff00ff) + (a >> 8 & 0x00ff00ff00ff00ff);
	a = (a & 0x0000ffff0000ffff) + (a >> 16 & 0x0000ffff0000ffff);
	return (a & 0xffffffff) + (a >> 32);
}
class unionfind {
	vector<int> par, rank, size_;//テゥツ?淌・ツコツヲテ」ツ?ァテ」ツ?ッテ」ツ?ェテ」ツ?湘」ツδ。テ」ツδ「テ」ツδェテ・ツ環ケテァツ篠?」ツつ津ィツ?ε」ツ?暗」ツつ凝」ツ?ェテ」ツつ詠ankテ」ツ?ョテ」ツ?凝」ツつ湘」ツつ甘」ツ?ォsizeテ」ツつ津、ツスツソテ」ツ??
public:
	unionfind(int n) :par(n), rank(n), size_(n, 1) {
		iota(ALL(par), 0);
	}
	int find(int x) {
		if (par[x] == x)return x;
		return par[x] = find(par[x]);
	}
	void unite(int x, int y) {
		x = find(x), y = find(y);
		if (x == y)return;
		if (rank[x] < rank[y])swap(x, y);
		par[y] = x;
		size_[x] += size_[y];
		if (rank[x] == rank[y])rank[x]++;
	}
	bool same(int x, int y) {
		return find(x) == find(y);
	}
	int size(int x) {
		return size_[find(x)];
	}
};
template<class T, class Func = function<T(T, T)>>
class segtree {
	vector<T> obj;
	int offset;
	Func updater;
	T e;
	int bufsize(int num) {
		int i = 1;
		for (; num >i; i <<= 1);
		offset = i - 1;
		return (i << 1) - 1;
	}
	T query(int a, int b, int k, int l, int r) {
		if (r <= a || b <= l)return e;
		if (a <= l && r <= b)return obj[k];
		else return updater(query(a, b, k * 2 + 1, l, (l + r) / 2), query(a, b, k * 2 + 2, (l + r) / 2, r));
	}
	/*
	template<class QF = function<T(T, T)>>
	auto query(int a, int b, int k, int l, int r,QF qf) {
		if (r <= a || b <= l)return e;
		if (a <= l && r <= b)return obj[k];
		else return qf(query(a, b, k * 2 + 1, l, (l + r) / 2, qf), query(a, b, k * 2 + 2, (l + r) / 2, r, qf));
	}*/
public:
	T query(int a, int b) {//[a,b)
		return query(a, b, 0, 0, offset + 1);
	}/*
	template<class QF = function<T(T, T)>>
	auto query(int a, int b,QF &&qf) {//[a,b)
		return query(a, b, 0, 0, offset + 1,qf);
	}*/
	void updateall(int l = 0, int r = -1) {
		if (r < 0)r = offset + 1;
		l += offset, r += offset;
		do {
			l = l - 1 >> 1, r = r - 1 >> 1;
			for (int i = l; i < r; i++)obj[i] = updater(obj[i * 2 + 1], obj[i * 2 + 2]);
		} while (l);
	}
	void update(int k, T &a) {
		k += offset;
		obj[k] = a;
		while (k) {
			k = k - 1 >> 1;
			obj[k] = updater(obj[k * 2 + 1], obj[k * 2 + 2]);
		}
	}
	segtree(int n, T e, const Func &updater = Func()) :obj(bufsize(n), e), e(e), updater(updater) {}
	segtree(vector<T> &vec, T e, const Func &updater = Func()) :obj(bufsize(vec.size()), e), e(e), updater(updater) {
		copy(vec.begin(), vec.end(), obj.begin() + offset);
		updateall();
	}
	typename vector<T>::reference operator[](int n) {
		return obj[n + offset];
	}
};
template<class T = int>
class BIT {//テ・ツ、ツ堙ヲツャツ。テ・ツ?ィITテ」ツ?ッテ」ツδォテ」ツδシテ」ツδ療」ツつ津」ツδ催」ツつケテ」ツδ暗」ツ?凖」ツつ古」ツ?ーテ」ツ??」ツ??」ツつ嘉」ツ?療」ツ??」ツ??
	vector<T> bit;
	int n;
public:
	BIT(int n) :bit(n + 1, 0) {}
	void add(int i, T x) {
		i++;
		while (i <= n) {
			bit[i] += x;
			i += i&-i;
		}
	}
	T sum(int i) {
		T s = 0;
		i++;
		while (i) {
			s += bit[i];
			i &= i - 1;
		}
		return s;
	}
};
template<class T = int>
class rangeadd {
	BIT<T> b0, b1;
public:
	rangeadd(int n) :b0(n), b1(n) {};
	void add(int l, int r, T x) {//[l,r)
		b0.add(l, -x*(l - 1));
		b1.add(l, x);
		b0.add(r, x*r);
		b1.add(r, -x);
	}
	T sum(int i) {
		if (i < 0)return 0;
		return b0.sum(i) + b1.sum(i)*i;
	}
	T sum(int l,int r) {
		return sum(r - 1) - sum(l - 1);
	}
};
//end of lib
//template<class S=void,int ptr_num, class T = char>class trie {
//	umap<T, trie<S, ptr_num, T> next;
//public:
//	S key;
//	trie<S, ptr_num, T>* ptr[ptr_num] = {};
//	trie(S &&data) :key(data) {}
//	trie(const S &data) :key(data) {}
//	void add(T x,S data) {
//		if (!next.find(x))next.insert(x, data);
//	}
//	trie& operator[](T x) {
//		return next[x];
//	}
//	bool find(T x) {
//		retun next.find(x);
//	}
//};
//template<class T=char>class AhoCorasick {
//	trie<pair<bool,int>, 2, T> tree;
//	AhoCorasick(vector<string> p) {
//		int num = 0;
//		vector<decltype(&tree)> que(p.size(),&tree);
//		for (int i = 0;; i++) {
//			bool end = 1;
//			int i = 0;
//			for (auto a : p) {
//				if (i >= a.size())break;
//				end = ;0
//				que[i] = (*que[i])[a[i]];
//				i++;
//			}
//			if (end)break;
//		}
//	}
//};
int main() {
	int n, m, k;
	cin >> n >> m >> k;
	vi d(m);
	rep(i, m)cin >> d[i];
	rep(i, m)d[i]--;
	vector<vi> v(n, vi(k));
	rep(i, n)rep(j, k)cin >> v[i][j];
	rep(i, n)rep(j, k)v[i][j]--;
	vector<vi> v2(1 << m, vi(k));
	vi m2(n);
	rep(i, m)m2[d[i]] = i + 1;
	rep(i, 1 << m) {
		rep(j, k) {
			int w = (1 << m) - 1;
			rep(k, m) {
				if (~i & (1 << k))w &= m2[v[d[k]][j]] ? ~(1 << m2[v[d[k]][j]] - 1) : -1;
			}
			v2[i][j] = w;
		}
	}
	heap<pii> h;
	h.push({ 0,0 });
	vi c(1 << m);
	while (h.top().second != (1 << m) - 1) {
		pii a = h.top(); h.pop();
		if (c[a.second])continue;
		c[a.second] = 1;
		rep(i, k)h.push({ a.first + 1,v2[a.second][i] });
	}
	cout << h.top().first << endl;
}