"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.util.HashSet;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.io.BufferedReader;
import java.util.Map;
import java.math.BigInteger;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.Set;
import java.util.StringTokenizer;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author shu_mj @ http://shu-mj.com
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskE solver = new TaskE();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskE {
    Scanner in;
    PrintWriter out;

    public void solve(int testNumber, Scanner in, PrintWriter out) {
        this.in = in;
        this.out = out;
        run();
    }

    void run() {
        int n = in.nextInt();
        int m = in.nextInt();
        Set<String> all = new HashSet<String>();
        Set<String> finish = new HashSet<String>();
        all.add("");
        int maxLength = 1;
        for (int i = 0; i < m; i++) {
            String s = in.next();
            maxLength = Math.max(maxLength, s.length());
            finish.add(s);
            for (int j = 0; j < s.length(); j++) {
                all.add(s.substring(0, j + 1));
            }
        }
        int last = 0;
        Map<String, Integer> index = new HashMap<String, Integer>();
        for (String s : all) {
            index.put(s, last++);
        }
        int[] isFinish = new int[last];
        for (String s : all) {
            for (String t : finish) {
                if (s.endsWith(t)) {
                    isFinish[index.get(s)] = Math.max(isFinish[index.get(s)], t.length());
                }
            }
        }
        int[][] next = new int[last][4];
        for (String s : all) {
            int start = index.get(s);
            for (int ch = 0; ch < 4; ch++) {
                String t = s + "AGCT".charAt(ch);
                while (!all.contains(t)) {
                    t = t.substring(1);
                }
                next[start][ch] = index.get(t);
            }
        }
        int[][] cnt = new int[last][maxLength];
        cnt[index.get("")][0] = 1;
        for (int len = 0; len < n; len++) {
            cnt = oneStep(cnt, maxLength, last, isFinish, next);
        }
        int res = 0;
        for (int state = 0; state < last; state++) {
            res += cnt[state][0];
            if (res >= M) res -= M;
        }
        out.println(res);
    }

    private int[][] oneStep(int[][] cnt, int maxLength, int last, int[] finish, int[][] next) {
        int[][] newCnt = new int[last][maxLength];
        for (int oldState = 0; oldState < last; 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;
                    newCnt[newState][newNeed] += cnt[oldState][oldNeed];
                    if (newCnt[newState][newNeed] >= M) newCnt[newState][newNeed] -= M;
                }
            }
        }
        return newCnt;
    }

    int M = 1000000009;
}

class Scanner {
    BufferedReader br;
    StringTokenizer st;

    public Scanner(InputStream in) {
        br = new BufferedReader(new InputStreamReader(in));
        eat("");
    }

    private void eat(String s) {
        st = new StringTokenizer(s);
    }

    public String nextLine() {
        try {
            return br.readLine();
        } catch (IOException e) {
            return null;
        }
    }

    public boolean hasNext() {
        while (!st.hasMoreTokens()) {
            String s = nextLine();
            if (s == null)
                return false;
            eat(s);
        }
        return true;
    }

    public String next() {
        hasNext();
        return st.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

}

