"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>
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:60000000")
using namespace std;
template <class T>
T sqr(T x) {
  return x * x;
}
double const pi = 3.1415926535897932384626433832795;
long long const inf64 = (long long)4e18;
const string name = "c";
const int NMAX = 1024;
const int PMAX = 555;
const int P = 1000000009;
const string ZN[5] = {"A", "C", "G", "T", "@"};
int n, m, d[NMAX][PMAX][11];
vector<string> pre, s;
int l[PMAX][5], r[PMAX][5], nxt[PMAX][5], used[13];
void calc(int i, int c) {
  string now = pre[i] + ZN[c];
  for (int j = (int)now.length() + 1 - 1; j >= 0; j--)
    for (int f = 0; f < (int)pre.size(); f++)
      if (now.substr(now.length() - j, j) == pre[f]) {
        nxt[i][c] = f;
        memset(used, 0, sizeof(used));
        for (int p = 0; p < (int)now.length() - j; p++)
          for (int t = 0; t < (int)m; t++)
            if (now.substr(p, s[t].length()) == s[t])
              for (int w = 0; w < (int)s[t].length(); w++) used[p + w] = 1;
        l[i][c] = r[i][c] = -1;
        for (int p = (int)12 - 1; p >= 0; p--)
          if (used[p]) {
            r[i][c] = p;
            while (p > 0 && used[p - 1]) p--;
            l[i][c] = p;
            break;
          }
        return;
      }
}
int main() {
  cin >> n >> m;
  pre.push_back("");
  for (int i = 0; i < (int)m; i++) {
    char buf[11];
    scanf("%s", buf);
    s.push_back((string)buf);
    for (int j = 0; j < (int)s[i].length(); j++)
      pre.push_back(s[i].substr(0, j + 1));
  }
  sort((pre).begin(), (pre).end());
  pre.resize(unique((pre).begin(), (pre).end()) - pre.begin());
  for (int i = 0; i < (int)pre.size(); i++)
    for (int c = 0; c < (int)5; c++) calc(i, c);
  memset(d, 0, sizeof(d));
  d[0][0][0] = 1;
  for (int i = 0; i < (int)n + 1; i++)
    for (int j = 0; j < (int)pre.size(); j++)
      for (int f = 0; f < (int)pre[j].size() + 1; f++)
        if (d[i][j][f] > 0)
          for (int c = 0; c < (int)5; c++)
            if ((c < 4 && i < n) || (c == 4 && i == n)) {
              int jj = nxt[j][c], ff = f + 1;
              if ((int)pre[j].size() - f >= l[j][c])
                ff = min(ff, (int)pre[j].length() - r[j][c]);
              if (ff <= (int)pre[jj].size() || i == n)
                d[i + 1][jj][ff] = (d[i + 1][jj][ff] + d[i][j][f]) % P;
            }
  cout << d[n + 1][0][1] << endl;
  return 0;
}
