"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;
template <typename T>
inline bool checkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool checkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int MOD = 1000000009;
int tot;
char str[100];
struct node {
  int len;
  node *next[4], *fail;
  node *clear() {
    for (int i = 0; i < 4; ++i) {
      next[i] = NULL;
    }
    fail = NULL;
    len = 0;
    return this;
  }
} tb[210], *root, *q[10000];
inline int change(char c) {
  if (c == 'A') return 0;
  if (c == 'T') return 1;
  if (c == 'G') return 2;
  return 3;
}
void insert(node *rt, char s[], int len) {
  node *p = rt;
  while (*s) {
    int i = change(*s);
    if (!p->next[i]) {
      p->next[i] = tb[tot++].clear();
    }
    p = p->next[i];
    s++;
  }
  p->len = len;
}
void build(node *rt) {
  q[0] = rt;
  for (int head = 0, tail = 1; head != tail; ++head) {
    node *h = q[head];
    for (int i = 0; i < 4; ++i) {
      node *p = h->fail;
      while (p && !p->next[i]) p = p->fail;
      if (h->next[i]) {
        if (p) {
          h->next[i]->fail = p->next[i];
          checkmax(h->next[i]->len, p->next[i]->len);
        } else
          h->next[i]->fail = rt;
        q[tail++] = h->next[i];
      } else if (p)
        h->next[i] = p->next[i];
      else
        h->next[i] = rt;
    }
  }
}
const int MAXLEN = 12;
int dp[1010][210][MAXLEN];
int main() {
  int N, M;
  while (scanf("%d%d", &N, &M) != EOF) {
    tot = 0;
    root = tb[tot++].clear();
    for (int i = 0; i < (M); ++i) {
      scanf("%s", str);
      insert(root, str, strlen(str));
    }
    build(root);
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    for (int i = 0; i < (N); ++i) {
      for (int node = 0; node < (tot); ++node) {
        for (int len = 0; len < (MAXLEN); ++len) {
          if (dp[i][node][len] == 0) continue;
          for (int j = 0; j < (4); ++j) {
            if (tb[node].next[j]->len > len) {
              dp[i + 1][tb[node].next[j] - tb][0] += dp[i][node][len];
              dp[i + 1][tb[node].next[j] - tb][0] %= MOD;
            } else if (len + 1 < MAXLEN) {
              dp[i + 1][tb[node].next[j] - tb][len + 1] += dp[i][node][len];
              dp[i + 1][tb[node].next[j] - tb][len + 1] %= MOD;
            }
          }
        }
      }
    }
    int ans = 0;
    for (int node = 0; node < (tot); ++node) {
      ans += dp[N][node][0];
      ans %= MOD;
    }
    printf("%d\n", ans);
  }
  return 0;
}
