"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.*;
import java.io.*;
import java.math.*;

public class Solution {
    final String ALPH = "ACGT";
    
    int n, m;
    String[] s;
    int suffCnt = 0;
    String[] suff = new String[500];
    Map<String, Integer> suffId = new HashMap<String, Integer>();
    
    public void go() throws Exception {
        Scanner sc = new Scanner(new InputStreamReader(System.in));
        n = sc.nextInt();
        m = sc.nextInt();
        s = new String[m];
        for (int i=0; i < s.length; i++)
            s[i] = sc.next();
        sc.close();
        
        for (int i=0; i < m; i++)
            for (int j=0; j <= s[i].length(); j++) {
                String pref = s[i].substring(0, j);
                if (!suffId.containsKey(pref)) {
                    suff[suffCnt] = pref;
                    suffId.put(pref, suffCnt++);
                }
            }
        
        int[][] nextSuff = new int[suffCnt][4];
        for (int i=0; i < suffCnt; i++)
            for (int j=0; j < 4; j++) {
                String have = suff[i] + (char)ALPH.charAt(j);
                while (!suffId.containsKey(have))
                    have = have.substring(1);
                nextSuff[i][j] = suffId.get(have);
            }
        
        int[] maxComplete = new int[suffCnt];
        for (int i=0; i < suffCnt; i++) {
            maxComplete[i] = -1;
            for (int j=0; j < m; j++)
                if (suff[i].endsWith(s[j]))
                    maxComplete[i] = Math.max(maxComplete[i], s[j].length());
        }
        
        int[][] cur = new int[suffCnt][11];
        cur[suffId.get("")][0] = 1;
        
        final int MOD = 1000000009;
        
        for (int pos = 1; pos <= n; pos++) {
            int[][] next = new int[suffCnt][11];
            for (int i=0; i < suffCnt; i++)
                for (int j=0; j <= 10; j++)
                    for (int k=0; k < 4; k++) {
                        int ni = nextSuff[i][k];
                        int nj = j+1;
                        if (maxComplete[ni] >= j+1)
                            nj = 0;
                        if (nj == 11) continue;
                        next[ni][nj] += cur[i][j];
                        while (next[ni][nj] >= MOD) next[ni][nj] -= MOD;
                    }
            for (int i=0; i < next.length; i++)
                for (int j=0; j < next[i].length; j++)
                    cur[i][j] = next[i][j];
        }
        
        int res = 0;
        for (int i=0; i < suffCnt; i++) {
            res += cur[i][0];
            while (res >= MOD) res -= MOD;
        }
        
        System.out.println(res);
    }
    
    public static void main(String[] args) throws Exception {
        (new Solution()).go();
    }
}
