"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.io.StreamTokenizer;
import java.util.Arrays;
import java.util.TreeSet;

public class F {
    
    static StreamTokenizer in;
    static PrintWriter out;
    
    static int nextInt() throws IOException {
        in.nextToken();
        return (int)in.nval;
    }
    
    static String nextString() throws IOException {
        in.nextToken();
        return in.sval;
    }

    public static void main(String[] args) throws IOException {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        
        int n = nextInt(), m = nextInt();
        String[] s = new String[m];
        for (int i = 0; i < m; i++) {
            s[i] = nextString();
        }
        
        Arrays.sort(s);
        
        TreeSet<String> set = new TreeSet<String>();
        set.add("");
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < s[i].length(); j++) {
                set.add(s[i].substring(0, j + 1));
            }
        }
        
        int q = set.size();
        String[] p = new String[q];
        int r = 0;
        for (String ss : set) {
            p[r++] = ss;
        }
        int[] end = new int[q];
        for (int i = 0; i < q; i++) {
            for (int j = 0; j < p[i].length(); j++) {
                if (Arrays.binarySearch(s, p[i].substring(j)) >= 0) {
                    end[i] = p[i].length() - j;
                    break;
                }
            }
        }
        
        int[][] v = new int[q][C];
        for (int i = 0; i < q; i++) {
            for (int j = 0; j < C; j++) {
                String t = p[i] + (char)(j+'A');
                while (!set.contains(t)) {
                    t = t.substring(1);
                }
                v[i][j] = Arrays.binarySearch(p, t);
            }
        }
        
        int[][][] dp = new int[n+1][L+1][q];
        dp[0][0][Arrays.binarySearch(p, "")] = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= L; j++) {
                for (int k = 0; k < q; k++) {
                    for (int x = 0; x < C; x++) {
                        int k2 = v[k][x];
                        if (j+1 <= p[k2].length()) {
                            if (j+1 <= end[k2]) {
                                dp[i+1][0][k2] += dp[i][j][k];
                                dp[i+1][0][k2] %= MD;
                            } else {
                                dp[i+1][j+1][k2] += dp[i][j][k];
                                dp[i+1][j+1][k2] %= MD;
                            }
                        }
                    }
                }
            }
        }
        
        int res = 0;
        for (int i = 0; i < q; i++) res = (res + dp[n][0][i])%MD;
        out.println(res);
        
        out.flush();
    }
    
    static int L = 20;
    static int C = 26;
    static int MD = 1000000009;
    
}