"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;
inline int read() {
  static char ch;
  bool sgn = false;
  while (ch = getchar(), ch < '0' || ch > '9')
    if (ch == '-') sgn = true;
  int res = ch - 48;
  while (ch = getchar(), ch >= '0' && ch <= '9') res = res * 10 + ch - 48;
  return sgn ? -res : res;
}
const int N = 1001;
const int Mod = 1e9 + 9;
int n, m, sind, ans, dp[2][N][101];
char str[10][N];
queue<int> Q;
struct node {
  int nxt[4];
  int fail, dep;
  void init() {
    memset(nxt, -1, sizeof(nxt));
    fail = dep = 0;
  }
} s[N];
int getnum(char a) {
  if (a == 'A') return 0;
  if (a == 'G') return 1;
  if (a == 'C') return 2;
  if (a == 'T') return 3;
}
void ins(char *str) {
  int len = strlen(str);
  int i, j, ind;
  for (i = ind = 0; i < len; i++) {
    j = getnum(str[i]);
    if (s[ind].nxt[j] == -1) {
      sind++;
      s[sind].init();
      s[ind].nxt[j] = sind;
    }
    ind = s[ind].nxt[j];
  }
  s[ind].dep = len;
}
void make_fail() {
  int i, ind, ind_f;
  while (!Q.empty()) Q.pop();
  for (i = 0; i < 4; i++) {
    if (s[0].nxt[i] != -1) {
      Q.push(s[0].nxt[i]);
    }
  }
  while (!Q.empty()) {
    ind = Q.front();
    Q.pop();
    for (int i = 0; i < 4; i++) {
      if (s[ind].nxt[i] != -1) {
        Q.push(s[ind].nxt[i]);
        ind_f = s[ind].fail;
        while (ind_f > 0 && s[ind_f].nxt[i] == -1) ind_f = s[ind_f].fail;
        if (s[ind_f].nxt[i] != -1) ind_f = s[ind_f].nxt[i];
        s[s[ind].nxt[i]].fail = ind_f;
        s[s[ind].nxt[i]].dep = max(s[ind_f].dep, s[s[ind].nxt[i]].dep);
      }
    }
  }
}
int main() {
  n = read();
  m = read();
  sind = 0;
  s[sind].init();
  for (int i = 0; i < m; i++) {
    scanf("%s", str[i]);
    ins(str[i]);
  }
  make_fail();
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    memset(dp[(i + 1) & 1], 0, sizeof(dp[(i + 1) & 1]));
    for (int l = max(0, i - 10); l <= i; l++) {
      for (int j = 0; j <= sind; j++) {
        if (dp[i & 1][l][j] == 0) continue;
        for (int k = 0; k < 4; k++) {
          int now = j;
          while (s[now].nxt[k] == -1 && now) now = s[now].fail;
          if (s[now].nxt[k] != -1) {
            now = s[now].nxt[k];
            if (i + 1 - l > s[now].dep)
              (dp[(i + 1) & 1][l][now] += dp[i & 1][l][j]) %= Mod;
            else
              (dp[(i + 1) & 1][i + 1][now] += dp[i & 1][l][j]) %= Mod;
          }
        }
      }
    }
  }
  for (int i = 0; i <= sind; i++) (ans += dp[n & 1][n][i]) %= Mod;
  cout << ans;
}
