"Multidimensional spaces are completely out of style these days, unlike genetics problems" — thought physicist Woll and changed his subject of study to bioinformatics. Analysing results of sequencing he faced the following problem concerning DNA sequences. We will further think of a DNA sequence as an arbitrary string of uppercase letters "A", "C", "G" and "T" (of course, this is a simplified interpretation).

Let w be a long DNA sequence and s1, s2, ..., sm — collection of short DNA sequences. Let us say that the collection filters w iff w can be covered with the sequences from the collection. Certainly, substrings corresponding to the different positions of the string may intersect or even cover each other. More formally: denote by |w| the length of w, let symbols of w be numbered from 1 to |w|. Then for each position i in w there exist pair of indices l, r (1 ≤ l ≤ i ≤ r ≤ |w|) such that the substring w[l ... r] equals one of the elements s1, s2, ..., sm of the collection.

Woll wants to calculate the number of DNA sequences of a given length filtered by a given collection, but he doesn't know how to deal with it. Help him! Your task is to find the number of different DNA sequences of length n filtered by the collection {si}.

Answer may appear very large, so output it modulo 1000000009.

Input

First line contains two integer numbers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10) — the length of the string and the number of sequences in the collection correspondently. 

Next m lines contain the collection sequences si, one per line. Each si is a nonempty string of length not greater than 10. All the strings consist of uppercase letters "A", "C", "G", "T". The collection may contain identical strings.

Output

Output should contain a single integer — the number of strings filtered by the collection modulo 1000000009 (109 + 9).

Examples

Input

2 1
A


Output

1


Input

6 2
CAT
TACT


Output

2

Note

In the first sample, a string has to be filtered by "A". Clearly, there is only one such string: "AA".

In the second sample, there exist exactly two different strings satisfying the condition (see the pictures below).

<image> <image>

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;

public class Main{
	
	String INPUT = "./data/judge/201709/C086C.txt";
	
	public static void main(String[] args) throws IOException {
		new Main().run();
	}
	
	static final int MOD = 1000000009;
	
	int solve(int n, String[] p) {
		int m = p.length;
		
		//���ÿ��ģʽ�ĺ�׺
		Set<String> pfx = new HashSet<>();
		pfx.add("");
		int maxLength = 1;
		for (int i = 0; i < m; ++i) {
			maxLength = Math.max(maxLength, p[i].length());
			for (int j = 1; j <= p[i].length(); ++j) {
				pfx.add(p[i].substring(0, j));
			}
		}
		
		int state = pfx.size();
		Map<String, Integer> mem = new HashMap<>();
		int idx = 0;
		for (String pf : pfx) {
			mem.put(pf, idx++);
		}
		
		//����ÿ��״̬��Ӧ������Ӵ�����
		int[] finish = new int[state];
		for (String pf : pfx) {
			for (String pp : p) {
				if (pf.endsWith(pp)) {
					finish[mem.get(pf)] = Math.max(finish[mem.get(pf)], pp.length());
				}
			}
		}
		
		//״̬ת��
		int[][] next = new int[state][4];
		for (String pf : pfx) {
			for (int ch = 0; ch < 4; ++ch) {
				String nxt = pf + "ATCG".charAt(ch);
				while (!mem.containsKey(nxt)) {
					nxt = nxt.substring(1);
				}
				next[mem.get(pf)][ch] = mem.get(nxt);
			}
		}
		
		int[][] dp = new int[state + 16][maxLength + 16];
		dp[mem.get("")][0] = 1;
		for (int i = 1; i <= n; ++i) {
			dp = oneStep(dp, maxLength, state, next, finish);
		}
		
		int ans = 0;
		for (int i = 0; i < state; ++i) {
			ans += dp[i][0];
			if (ans >= MOD) ans -= MOD;
		}
		return ans;
	}
	
	int[][] oneStep(int[][] cnt, int maxLength, int state, int[][] next, int[] finish) {
		int[][] newDp = new int[state + 16][maxLength + 16];
		for (int oldState = 0; oldState < state; ++oldState) {
			for (int oldNeed = 0; oldNeed < maxLength; ++oldNeed) {
				for (int ch = 0; ch < 4; ++ch) {
					int newState = next[oldState][ch];
					int newNeed  = oldNeed + 1;
					if (newNeed <= finish[newState]) newNeed = 0;
					if (newNeed >= maxLength) continue;
					newDp[newState][newNeed] += cnt[oldState][oldNeed];
					if (newDp[newState][newNeed] >= MOD)  newDp[newState][newNeed] -= MOD;
				}
			}
		}
		return newDp;
	}
	
	void read() {
		int n = ni();
		int m = ni();
		String[] p = new String[m];
		for (int i = 0; i < m; ++i) {
			p[i] = ns();
		}
		out.println(solve(n, p));
	}

	FastScanner in;
	PrintWriter out;
	
	void run() throws IOException {
		boolean oj;
		try {
			oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
		} catch (Exception e) {
			oj = System.getProperty("ONLINE_JUDGE") != null;
		}
		
		InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
		in = new FastScanner(is);
		out = new PrintWriter(System.out);
		long s = System.currentTimeMillis();
		read();
		out.flush();
		if (!oj){
			System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
		}
	}
	
	public boolean more(){
		return in.hasNext();
	}
	
	public int ni(){
		return in.nextInt();
	}
	
	public long nl(){
		return in.nextLong();
	}
	
	public double nd(){
		return in.nextDouble();
	}
	
	public String ns(){
		return in.nextString();
	}
	
	public char nc(){
		return in.nextChar();
	}
	
	class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		boolean hasNext;

		public FastScanner(InputStream is) throws IOException {
			br = new BufferedReader(new InputStreamReader(is));
			hasNext = true;
		}

		public String nextToken() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					hasNext = false;
					return "##";
				}
			}
			return st.nextToken();
		}
		
		String next = null;
		public boolean hasNext(){
			next = nextToken();
			return hasNext;
		}

		public int nextInt() {
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return Integer.parseInt(more);
		}

		public long nextLong() {
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return Long.parseLong(more);
		}

		public double nextDouble() {
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return Double.parseDouble(more);
		}
		
		public String nextString(){
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return more;
		}
		
		public char nextChar(){
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return more.charAt(0);
		}
	}
	
	static class ArrayUtils {

		public static void fill(int[][] f, int value) {
			for (int i = 0; i < f.length; ++i) {
				Arrays.fill(f[i], value);
			}
		}
		
		public static void fill(int[][][] f, int value) {
			for (int i = 0; i < f.length; ++i) {
				fill(f[i], value);
			}
		}
		
		public static void fill(int[][][][] f, int value) {
			for (int i = 0; i < f.length; ++i) {
				fill(f[i], value);
			}
		}
	}
}


