"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 = 150;
const int Mod = 1e9 + 9;
int n, m, ans, tot, q[N], val[N], fail[N], c[N][4], f[2][N][12];
char s[20];
template <class T>
inline void in(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9';) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  for (; ch >= '0' && ch <= '9';) x = x * 10 + ch - '0', ch = getchar();
  x = f ? -x : x;
}
inline int trs(char ch) {
  return ch == 'A' ? 0 : (ch == 'C' ? 1 : (ch == 'G') ? 2 : 3);
}
inline void ins() {
  int x = 1, ls = strlen(s + 1);
  for (int w, i = 1; i <= ls; ++i) {
    w = trs(s[i]);
    if (!c[x][w]) c[x][w] = ++tot;
    x = c[x][w];
  }
  val[x] = ls;
}
inline void build() {
  int x, u, v, l = 0, r = 0;
  for (int i = 0; i < 4; ++i)
    if (c[1][i])
      q[++r] = c[1][i], fail[c[1][i]] = 1;
    else
      c[1][i] = 1;
  for (; l < r;) {
    u = q[++l], val[u] = max(val[u], val[fail[u]]);
    for (int i = 0; i < 4; ++i)
      if ((v = c[u][i]))
        fail[v] = c[fail[u]][i], q[++r] = v;
      else
        c[u][i] = c[fail[u]][i];
  }
}
inline void solve() {
  f[0][1][0] = 1;
  int v, now = 0, lst = 1;
  for (int t = 1; t <= n; ++t) {
    now ^= 1, lst ^= 1;
    memset(f[now], 0, sizeof f[now]);
    for (int j = 1; j <= tot; ++j)
      for (int k = 0; k <= 10; ++k)
        if (f[lst][j][k])
          for (int i = 0; i < 4; ++i)
            if ((v = c[j][i])) {
              if (val[v] > k)
                (f[now][v][0] += f[lst][j][k]) %= Mod;
              else
                (f[now][v][k + 1] += f[lst][j][k]) %= Mod;
            }
  }
  for (int i = 1; i <= tot; ++i) ans = (ans + f[now][i][0]) % Mod;
  printf("%d\n", ans);
}
int main() {
  in(n), in(m), tot = 1;
  for (int i = 1; i <= m; ++i) scanf("%s", s + 1), ins();
  build(), solve();
  return 0;
}
