"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;
int mod = 1000000009;
char str[200][200];
int nwnode[104][5], backnode[104], cnt[104], id = 0, level[104], len;
int casio = 1;
int name[500];
int dp[103][1003][20], my[120];
vector<int> vc[120];
int newnode() {
  id++;
  for (int i = 1; i <= 4; i++) {
    nwnode[id][i] = 0;
  }
  cnt[id] = 0;
  return id;
}
void build(int n) {
  int root = newnode(), p;
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    p = root;
    level[root] = 0;
    for (int j = 0; str[i][j]; j++) {
      int c = name[str[i][j]];
      if (!nwnode[p][c]) nwnode[p][c] = newnode();
      level[nwnode[p][c]] = level[p] + 1;
      p = nwnode[p][c];
    }
    cnt[p] = 1;
  }
  for (int i = 1; i <= 4; i++) {
    if (!nwnode[root][i])
      nwnode[root][i] = root;
    else {
      q.push(nwnode[root][i]);
      backnode[nwnode[root][i]] = 1;
      vc[nwnode[1][i]].push_back(1);
    }
  }
  int u, v, w, c;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (int i = 1; i <= 4; i++) {
      if (!nwnode[u][i]) {
        nwnode[u][i] = nwnode[backnode[u]][i];
        continue;
      }
      w = backnode[u];
      v = nwnode[u][i];
      while (nwnode[w][i] == 0) {
        w = backnode[w];
      }
      int c = nwnode[w][i];
      backnode[v] = w = c;
      q.push(v);
      vc[v].push_back(w);
    }
  }
}
int n, mxlen;
int dp_func(int node, int len, int mx) {
  if (len == n + 1) {
    return mx == 1;
  }
  if (mx > mxlen) return 0;
  if (dp[node][len][mx] != -1) return dp[node][len][mx];
  long long ret = 0;
  for (int i = 1; i <= 4; i++) {
    int p = nwnode[node][i];
    if (p == 1) continue;
    if (level[p] < mx) continue;
    if (my[p] >= mx)
      ret += dp_func(p, len + 1, 1);
    else
      ret += dp_func(p, len + 1, mx + 1);
  }
  return dp[node][len][mx] = ret % mod;
}
int m_func(int v) {
  if (cnt[v]) return my[v] = level[v];
  if (my[v] != -1) return my[v];
  int ret = 0;
  for (int i = 0; i < vc[v].size(); i++) {
    int w = vc[v][i];
    ret = max(ret, m_func(w));
  }
  return my[v] = ret;
}
int main() {
  name['A'] = 1;
  name['C'] = 2;
  name['G'] = 3;
  name['T'] = 4;
  int i, j, k, l, m;
  memset(my, -1, sizeof my);
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", str[i]);
    mxlen = max(mxlen, (int)strlen(str[i]));
  }
  build(m);
  for (int i = 1; i <= id; i++) {
    m_func(i);
  }
  memset(dp, -1, sizeof dp);
  int ret = dp_func(1, 1, 1);
  cout << ret << endl;
}
