"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.util.Comparator;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Collection;
import java.util.List;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.TreeMap;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.math.BigInteger;
import java.util.Collections;
import java.io.InputStream;

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

class TaskC {
    public static final int MOD = 1000000009;

    TreeMap<String, Integer> allPrefixes;

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int m = in.nextInt();
        s = new String[m];
        allPrefixes = new TreeMap<String, Integer>();
        allPrefixes.put("", 0);
        for (int i = 0; i < m; ++i) {
            s[i] = in.nextString();
            for (int j = 0; j < s[i].length(); ++j) {
                String pref = s[i].substring(0, j + 1);
                if (!allPrefixes.containsKey(pref)) {
                    allPrefixes.put(pref, allPrefixes.size());
                }
            }
        }


        dp = new int[n+1][11][m * 12];
        ArrayUtils.fill(dp, -1);
        out.println(rec("", n, 0));
    }

    final String letters = "ACGT";

    private int rec(String s, int n, int notCovered) {
        if (n == 0) {
            return notCovered == 0 ? 1 : 0;
        }
        int id = allPrefixes.get(s);
        if (dp[n][notCovered][id] != -1) return dp[n][notCovered][id];
        int res = 0;
        for (int i = 0; i < 4; ++i) {
            char c = letters.charAt(i);
            String nextPrefix = getNextPrefix(s, c);
            if (nextPrefix.length() == 0) continue;
            int newNotCovered = notCovered + 1;
            if (maxWordInPrefix(nextPrefix) >= newNotCovered) {
                newNotCovered = 0;
            }
            if (newNotCovered > nextPrefix.length()) {
                continue;
            }
            res += rec(nextPrefix, n - 1, newNotCovered);
            if (res >= MOD) res -= MOD;
        }
        dp[n][notCovered][id] = res;
        return res;
    }

    private int maxWordInPrefix(String t) {
        for (int i = t.length(); i >= 0; --i) {
            String sub = t.substring(t.length() - i);
            for (String ss : s) {
                if (ss.compareTo(sub) == 0) {
                    return i;
                }
            }
        }
        return 0;

    }

    private String getNextPrefix(String s, char c) {
        s += c;
        for (int i = s.length(); i >= 0; --i) {
            String sub = s.substring(s.length() - i);
            if (allPrefixes.containsKey(sub)) {
                return sub;
            }
        }
        return "";
    }

    String[] s;
    int[][][] dp;
}

class InputReader {
    private BufferedReader reader;
    private StringTokenizer stt;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
    }

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

    public String nextString() {
        while (stt == null || !stt.hasMoreTokens()) {
            stt = new StringTokenizer(nextLine());
        }
        return stt.nextToken();
    }

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

}

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[][][] a, int value) {
        for (int i = 0; i < a.length; ++i) {
            fill(a[i], value);
        }
    }

}

