"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>
const int oo = 2139063143;
const int N = 210;
const int P = 1000000009;
using namespace std;
inline void sc(int &x) {
  x = 0;
  static int p;
  p = 1;
  static char c;
  c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - 48);
    c = getchar();
  }
  x *= p;
}
inline void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
inline void pr(int x) { print(x), putchar('\n'); }
void add(int &x, int y) {
  x += y;
  x = x >= P ? x - P : x;
}
int change(char c) {
  if (c == 'A') return 1;
  if (c == 'G') return 2;
  if (c == 'C') return 3;
  return 4;
}
struct ACAM {
  int fa, son[5], len;
} t[N];
int totp = 1;
void ins(char *s, int n) {
  int p = 1;
  for (int i = 1; i <= n; i++) {
    int &v = t[p].son[change(s[i])];
    if (!v) v = ++totp;
    p = v;
  }
  t[p].len = max(t[p].len, n);
}
queue<int> q;
void build() {
  for (int i = 1; i <= 4; i++) t[0].son[i] = 1;
  q.push(1);
  while (q.size()) {
    int u = q.front(), f = t[u].fa;
    q.pop();
    for (int c = 1; c <= 4; c++) {
      int &v = t[u].son[c], g = t[f].son[c];
      if (v)
        t[v].fa = g, q.push(v);
      else
        v = g;
    }
    t[u].len = max(t[u].len, t[f].len);
  }
}
char s[N];
int f[1010][N][15];
int main() {
  int n, m, ma = 0;
  sc(n), sc(m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    ins(s, len);
    ma = max(ma, len);
  }
  build();
  f[0][1][0] = 1;
  for (int i = 0; i <= n; i++) {
    for (int u = 1; u <= totp; u++) {
      for (int j = 0; j < ma; j++) {
        int g = f[i][u][j];
        for (int c = 1; c <= 4; c++)
          if (g) {
            int v = t[u].son[c];
            if (t[v].len >= j + 1)
              add(f[i + 1][v][0], g);
            else
              add(f[i + 1][v][j + 1], g);
          }
      }
    }
  }
  int ans = 0;
  for (int u = 1; u <= totp; u++) add(ans, f[n][u][0]);
  pr(ans);
  return 0;
}
