"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 toii[500];
struct ac_node {
  int id, ch[4], fail;
} t[110];
int ac_node_cnt;
int dp[1010][110][15], len[15];
char s[15][15];
int pre[110];
inline int new_ac_node() {
  int x = ++ac_node_cnt;
  t[x].id = t[x].fail = 0;
  return memset(t[x].ch, 0, sizeof(t[x].ch)), x;
}
inline int insert_ac(int rt, char *s, int n, int id) {
  int x = rt;
  for (int i = 1; i <= n; i++) {
    int c = toii[(int)s[i]];
    if (!t[x].ch[c]) t[x].ch[c] = new_ac_node();
    x = t[x].ch[c];
  }
  t[x].id = id;
  return 0;
}
queue<int> q;
inline int get_fail(int rt) {
  while (!q.empty()) q.pop();
  for (int i = 0; i < 4; i++) {
    int &y = t[rt].ch[i];
    if (y)
      t[y].fail = rt, q.push(y), pre[y] = t[y].id;
    else
      y = rt;
  }
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      int &y = t[x].ch[i], f = t[x].fail, c = t[f].ch[i];
      if (y) {
        t[y].fail = c, q.push(y);
        if (t[y].id)
          pre[y] = t[y].id;
        else
          pre[y] = pre[c];
      } else
        y = c;
    }
  }
  return 0;
}
inline int get_dp(int n) {
  int m = ac_node_cnt, v;
  dp[0][1][0] = 1;
  for (int i = 0; i < n; i++)
    for (int x = 1; x <= m; x++)
      for (int j = 0; j < 10; j++)
        if ((v = dp[i][x][j]))
          for (int c = 0; c < 4; c++) {
            int y = t[x].ch[c];
            if (len[pre[y]] >= j + 1)
              (dp[i + 1][y][0] += v) %= 1000000009;
            else
              (dp[i + 1][y][j + 1] += v) %= 1000000009;
          }
  int ans = 0;
  for (int i = 1; i <= m; i++) (ans += dp[n][i][0]) %= 1000000009;
  return ans;
}
int main() {
  toii['A'] = 0, toii['T'] = 1, toii['C'] = 2, toii['G'] = 3;
  int n, m, rt;
  scanf("%d%d", &n, &m), rt = new_ac_node();
  for (int i = 1; i <= m; i++)
    scanf("%s", s[i] + 1), insert_ac(rt, s[i], len[i] = strlen(s[i] + 1), i);
  return get_fail(rt), !printf("%d\n", get_dp(n));
}
