"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 <class T>
inline void checkmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (a < b) a = b;
}
const int oo = 1 << 30;
const double eps = 1e-7;
const int N = 128;
const int M = 4;
const long long P = 1000000009;
struct tree {
  tree *next[M], *fail;
  int end;
};
tree vd;
tree tr[N];
int len;
int trn;
int to[N];
tree *Q[N];
int xn;
int dp[1024][10][N];
void add_tr() {
  char s[N];
  int i, j, k;
  int sn, tmp;
  scanf("%s", s);
  sn = strlen(s);
  tree *p = tr;
  for ((i) = 0; i < (sn); (i)++) {
    tmp = to[s[i]];
    if (NULL == p->next[tmp]) {
      tr[trn] = vd;
      p->next[tmp] = tr + (trn++);
    }
    p = p->next[tmp];
  }
  p->end = sn;
}
void bfs() {
  int i, j, k, f, b;
  tree *p, *q;
  p = tr;
  f = b = 0;
  for ((i) = 0; i < (M); (i)++)
    if (p->next[i]) {
      Q[b++] = p->next[i];
      p->next[i]->fail = tr;
    } else {
      p->next[i] = tr;
    }
  while (f < b) {
    p = Q[f++];
    checkmax(p->end, p->fail->end);
    for ((i) = 0; i < (M); (i)++)
      if (p->next[i]) {
        Q[b++] = p->next[i];
        p->next[i]->fail = p->fail->next[i];
      } else
        p->next[i] = p->fail->next[i];
  }
}
int read() {
  int n, m;
  if (2 != scanf("%d %d", &n, &m)) return 0;
  len = n;
  tr[0] = vd;
  trn = 1;
  while (m--) add_tr();
  xn = trn;
  bfs();
  return 1;
}
void pre() {
  (memset((to), 0, sizeof(to)));
  to['A'] = 0;
  to['C'] = 1;
  to['G'] = 2;
  to['T'] = 3;
  memset(&vd, 0, (sizeof(vd)));
}
int dfs(int len, int left, int id) {
  if (len < 0 || left > 9) return !left;
  int &r = dp[len][left][id];
  if (-1 != r) return r;
  r = 0;
  int i;
  tree *p = tr + id;
  for ((i) = 0; i < (M); (i)++) {
    r += dfs(len - 1, p->next[i]->end > left ? 0 : left + 1, p->next[i] - tr);
    if (r >= P) r -= P;
  }
  return r;
}
int doit() {
  (memset((dp), -1, sizeof(dp)));
  return dfs(len - 1, 0, 0);
}
int main() {
  int T, _ = 0;
  pre();
  while (read()) {
    int ans = doit();
    printf("%d\n", ans);
  }
  return 0;
}
