"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>
using namespace std;
const int inf = 1000 * 1000 * 1000;
const int NMAX = 200;
const long long MOD = 1000 * 1000 * 1000 + 9;
const int K = 27;
struct vertex {
  int next[K];
  bool leaf;
  int p;
  char pch;
  int link;
  int go[K];
  char covered;
  int len;
};
vertex t[NMAX + 1];
int sz;
long long Pows[1000];
void Init() {
  t[0].p = t[0].link = -1;
  memset(t[0].next, 255, sizeof t[0].next);
  memset(t[0].go, 255, sizeof t[0].go);
  t[0].covered = 0;
  sz = 1;
  t[0].len = 0;
}
void add_string(const string &s) {
  int v = 0;
  for (size_t i = 0; i < s.length(); ++i) {
    char c = s[i] - 'A';
    if (t[v].next[c] == -1) {
      memset(t[sz].next, 255, sizeof t[sz].next);
      memset(t[sz].go, 255, sizeof t[sz].go);
      t[sz].covered = -1;
      t[sz].link = -1;
      t[sz].p = v;
      t[sz].pch = c;
      t[sz].len = i + 1;
      t[v].next[c] = sz++;
    }
    v = t[v].next[c];
  }
  t[v].leaf = true;
}
int go(int v, char c);
int get_link(int v) {
  if (t[v].link == -1)
    if (v == 0 || t[v].p == 0)
      t[v].link = 0;
    else
      t[v].link = go(get_link(t[v].p), t[v].pch);
  return t[v].link;
}
int Covered(int v) {
  if (t[v].covered == -1) {
    if (t[v].leaf)
      t[v].covered = t[v].len;
    else
      t[v].covered = Covered(get_link(v));
  }
  return t[v].covered;
}
int go(int v, char c) {
  if (t[v].go[c] == -1)
    if (t[v].next[c] != -1)
      t[v].go[c] = t[v].next[c];
    else
      t[v].go[c] = v == 0 ? 0 : go(get_link(v), c);
  return t[v].go[c];
}
int go(int v, char c, char &err) {
  err = 0;
  if (t[v].next[c] != -1)
    t[v].go[c] = t[v].next[c];
  else {
    err = 1;
    t[v].go[c] = v == 0 ? 0 : go(get_link(v), c);
  }
  return t[v].go[c];
}
int n, m;
vector<string> v;
long long dp[300][1009][23];
long long DP(int pos, int ost, int free) {
  if (ost == 0 && (Covered(pos) >= free)) return 1;
  if (ost == 0) {
    return 0;
  }
  if (dp[pos][ost][free] != -1) return dp[pos][ost][free];
  long long res = 0;
  for (int i = 0; i <= 25; i++) {
    char err;
    int T = go(pos, i, err);
    if (!err)
      if (Covered(T) >= free + 1)
        res += DP(T, ost - 1, 0);
      else
        res += DP(T, ost - 1, free + 1);
    else {
      if (free + 1 > t[T].len)
        res = res;
      else {
        if (Covered(T) >= free + 1)
          res += DP(T, ost - 1, 0);
        else
          res += DP(T, ost - 1, free + 1);
      }
    }
    res %= MOD;
  }
  dp[pos][ost][free] = res;
  return res;
};
int main() {
  memset(dp, -1, sizeof(dp));
  ;
  Pows[0] = 1;
  for (int i = 1; i < 1000; i++) {
    Pows[i] = Pows[i - 1] * 26;
    Pows[i] %= MOD;
  }
  Init();
  scanf("%d%d", &n, &m);
  v.resize(m);
  for (int i = 0; i < m; i++) {
    cin >> v[i];
    if (v[i].size() <= n) add_string(v[i]);
  }
  long long res = Pows[n];
  res = DP(0, n, 0);
  res %= MOD;
  res += MOD;
  res %= MOD;
  cout << res;
  return 0;
}
