"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 MAXN = 3e5 + 10;
const double eps = 1e-8;
const int mod = 1e9 + 9;
using namespace std;
struct edge {
  int t, v;
  edge *next;
} e[MAXN << 1], *h[MAXN], *o = e;
void add(int x, int y, int vul) {
  o->t = y;
  o->v = vul;
  o->next = h[x];
  h[x] = o++;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int T[MAXN];
typedef struct node {
  int dep, ch[4], fair, pos;
  bool vis;
} node;
int rt, n, m, cnt;
node d[105];
void newnode(int &x, int y) {
  x = ++cnt;
  d[x].dep = d[y].dep + 1;
  d[x].fair = 0;
  d[x].pos = 0;
}
void insert(char str[]) {
  int len = strlen(str);
  int temp = rt;
  for (int i = 0; i < len; i++) {
    int t = T[str[i] - 'A'];
    if (!d[temp].ch[t]) newnode(d[temp].ch[t], temp);
    temp = d[temp].ch[t];
  }
  d[temp].vis = 1;
  d[temp].pos = temp;
}
queue<int> que;
void Ac() {
  que.push(rt);
  while (!que.empty()) {
    int x = que.front();
    que.pop();
    for (int i = 0; i < 4; i++) {
      if (!d[x].ch[i]) {
        d[x].ch[i] = d[d[x].fair].ch[i];
        continue;
      }
      if (!d[x].fair) {
        d[d[x].ch[i]].fair = rt;
        que.push(d[x].ch[i]);
        continue;
      }
      int temp = d[x].fair;
      while (temp && !d[temp].ch[i]) temp = d[temp].fair;
      if (!temp)
        d[d[x].ch[i]].fair = rt;
      else
        d[d[x].ch[i]].fair = d[temp].ch[i];
      if (d[d[d[x].ch[i]].fair].dep > d[d[d[x].ch[i]].pos].dep)
        d[d[x].ch[i]].pos = d[d[d[x].ch[i]].fair].pos;
      que.push(d[x].ch[i]);
    }
  }
}
int dp[1005][101][21];
char str[21];
int main() {
  T['A' - 'A'] = 0;
  T['C' - 'A'] = 1;
  T['G' - 'A'] = 2;
  T['T' - 'A'] = 3;
  n = read();
  m = read();
  newnode(rt, 0);
  d[rt].dep = 0;
  for (int i = 1; i <= m; i++) scanf("%s", str), insert(str);
  Ac();
  dp[0][1][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= cnt; j++) {
      for (int k = 0; k < 4; k++) {
        int pos = d[j].ch[k];
        if (!pos) continue;
        for (int p = 0; p <= 9; p++) {
          if (d[d[pos].pos].dep > p) {
            dp[i + 1][pos][0] += dp[i][j][p];
            dp[i + 1][pos][0] %= mod;
          } else {
            dp[i + 1][pos][p + 1] += dp[i][j][p];
            dp[i + 1][pos][p + 1] %= mod;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= cnt; i++) ans += dp[n][i][0], ans %= mod;
  printf("%lld\n", ans);
  return 0;
}
