"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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Genetic implements Runnable {
    private void solve() throws IOException {
        int n = nextInt();
        int m = 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 = nextToken();
            maxLength = Math.max(maxLength, s.length());
            finish.add(s);
            for (int j = 0; j < s.length(); ++j)
                all.add(s.substring(0, j + 1));
        }
        Map<String, Integer> index = new HashMap<String, Integer>();
        int last = 0;
        for (String st : all) {
            index.put(st, last++);
        }
        int[] isFinish = new int[last];
        int[][] next = new int[last][4];
        for (String s : all)
            for (String t : finish)
                if (s.endsWith(t))
                    isFinish[index.get(s)] = Math.max(isFinish[index.get(s)], t.length());
        for (String s : all) {
            int start = index.get(s);
            for (int ch = 0; ch < 4; ++ch) {
                String t = s + "ACGT".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 >= MODULO)
                res -= MODULO;
        }
        writer.println(res);
    }

    static final int MODULO = 1000000009;

    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] >= MODULO) {
                        newCnt[newState][newNeed] -= MODULO;
                    }
                }
            }
        return newCnt;
    }

    public static void main(String[] args) {
        new Genetic().run();
    }

    BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter writer;

    public void run() {
        try {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            writer = new PrintWriter(System.out);
            solve();
            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }
}
