"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 MAXN = 10050, mod = 1e9 + 9;
namespace AC {
int c[MAXN][5], len[MAXN], maxlen[MAXN], fail[MAXN], cnt, rt;
void ins(char *s) {
  int n = strlen(s);
  int now = rt;
  for (int i = 0; i < n; i++) {
    int a = s[i] - 'a';
    if (!c[now][a]) c[now][a] = ++cnt;
    now = c[now][a];
  }
  len[now] = max(len[now], n);
}
void getfail() {
  queue<int> q;
  for (int i = 0; i < 4; i++) {
    if (c[rt][i] != 0) {
      fail[c[rt][i]] = rt;
      q.push(c[rt][i]);
    }
  }
  while (!q.empty()) {
    int nown = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      int &v = c[nown][i];
      if (v == 0) {
        v = c[fail[nown]][i];
      } else {
        fail[v] = c[fail[nown]][i];
        len[v] = max(len[v], len[fail[v]]);
        q.push(v);
      }
    }
  }
}
}  // namespace AC
int n, m;
int dp[MAXN][105][12];
char s[MAXN];
int dfs(int len, int nownode, int nowleft) {
  if (nowleft > 10) return 0;
  if (dp[len][nownode][nowleft] != -1) {
    return dp[len][nownode][nowleft];
  }
  if (len == 0 && nowleft == 0)
    return 1;
  else if (len == 0)
    return 0;
  int ans = 0;
  for (int i = 0; i < 4; i++) {
    int v = AC::c[nownode][i];
    ans += dfs(len - 1, v, (AC::len[v] >= nowleft + 1) ? 0 : nowleft + 1);
    ans %= mod;
  }
  return dp[len][nownode][nowleft] = ans;
}
void init() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s);
    int t = strlen(s);
    for (int i = 0; i < t; i++) {
      if (s[i] == 'A') {
        s[i] = 'a';
      } else if (s[i] == 'C') {
        s[i] = 'b';
      } else if (s[i] == 'T') {
        s[i] = 'c';
      } else if (s[i] == 'G') {
        s[i] = 'd';
      } else {
        printf("-1\n");
      }
    }
    AC::ins(s);
  }
}
void solve() {
  AC::getfail();
  memset(dp, -1, sizeof(dp));
  printf("%d\n", dfs(n, 0, 0));
}
int main() {
  init();
  solve();
  return 0;
}
