"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 N = 105;
const int MOD = 1e9 + 9;
template <typename T>
inline void read(T &AKNOI) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  AKNOI = flag * x;
}
inline void Inc(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
char s[20];
int n, m, id[255];
int tot, ch[N][4], fail[N], ed[N], q[N];
int dp[2][N][10], ans;
void Insert() {
  int p = 0, len = strlen(s + 1);
  for (int i = 1; i <= len; ++i) {
    int x = id[s[i]];
    if (!ch[p][x]) {
      ch[p][x] = ++tot;
    }
    p = ch[p][x];
  }
  ed[p] = len;
}
void GetFail() {
  int head = 1, tail = 0;
  for (int i = 0; i < 4; ++i) {
    if (ch[0][i]) {
      q[++tail] = ch[0][i];
    }
  }
  while (head <= tail) {
    int u = q[head++];
    ed[u] = max(ed[u], ed[fail[u]]);
    for (int i = 0; i < 4; ++i) {
      if (ch[u][i]) {
        fail[ch[u][i]] = ch[fail[u]][i];
        q[++tail] = ch[u][i];
      } else {
        ch[u][i] = ch[fail[u]][i];
      }
    }
  }
}
void init() {
  id['A'] = 0;
  id['C'] = 1;
  id['G'] = 2;
  id['T'] = 3;
  read(n);
  read(m);
  for (int i = 1; i <= m; ++i) {
    scanf("%s", s + 1);
    Insert();
  }
  GetFail();
}
void solve() {
  dp[0][0][0] = 1;
  int pre = 0, cur = 1;
  for (int i = 1; i <= n; ++i) {
    memset(dp[cur], 0, sizeof(dp[cur]));
    for (int j = 0; j <= tot; ++j) {
      for (int k = 0; k < 10; ++k) {
        if (!dp[pre][j][k]) continue;
        for (int t = 0; t < 4; ++t) {
          int r = ch[j][t];
          if (ed[r] > k) {
            Inc(dp[cur][r][0], dp[pre][j][k]);
          } else if (k + 1 < 10) {
            Inc(dp[cur][r][k + 1], dp[pre][j][k]);
          }
        }
      }
    }
    swap(pre, cur);
  }
  for (int j = 0; j <= tot; ++j) {
    Inc(ans, dp[pre][j][0]);
  }
  printf("%d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}
