"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 INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 1007;
const int MAXS = 107;
const int MAXM = 13;
const int MAXL = 13;
const int BASE = 4;
const int MOD = 1e9 + 9;
int toid[128];
int n, m, kn, ml;
char s[MAXM][MAXL];
int nxt[MAXS][BASE], fail[MAXS], q[MAXS];
int ok[MAXS];
int tag[MAXS];
int dp[MAXN][MAXL][MAXS];
void init();
void input();
void work();
void trie_add(char s[]);
void buildac();
int main() {
  init();
  input();
  work();
}
void init() {
  ios::sync_with_stdio(false);
  toid['A'] = 0;
  toid['C'] = 1;
  toid['G'] = 2;
  toid['T'] = 3;
}
void input() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) scanf("%s", s[i]);
}
void work() {
  kn = 1;
  memset(nxt, -1, sizeof(nxt));
  memset(tag, 0, sizeof(tag));
  for (int i = 0; i < m; ++i)
    trie_add(s[i]), ml = (((ml) > (strlen(s[i]))) ? (ml) : (strlen(s[i])));
  buildac();
  ++ml;
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  int v, nj, nl;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < ml; ++j) {
      for (int k = 0; k < kn; ++k) {
        if (dp[i][j][k]) {
          for (int nc = 0; nc < BASE; ++nc) {
            v = nxt[k][nc];
            nj = j + 1;
            if (tag[v] || ok[v]) {
              if (tag[v])
                nl = tag[v];
              else
                nl = tag[ok[v]];
              if (nl >= nj) nj = 0;
            }
            dp[i + 1][nj][v] += dp[i][j][k];
            dp[i + 1][nj][v] %= MOD;
          }
        }
      }
    }
  }
  int sum = 0;
  for (int i = 0; i < kn; ++i) {
    sum += dp[n][0][i];
    sum %= MOD;
  }
  printf("%d\n", sum);
}
void trie_add(char s[]) {
  int len = strlen(s), now = 0, nc;
  for (int i = 0; i < len; ++i) {
    nc = toid[s[i]];
    if (nxt[now][nc] ^ -1)
      now = nxt[now][nc];
    else
      now = nxt[now][nc] = kn++;
  }
  tag[now] = len;
}
void buildac() {
  int l = 0, r = 0, now, v, f;
  q[r++] = 0;
  fail[0] = 0;
  while (l < r) {
    now = q[l++];
    for (int i = 0; i < BASE; ++i) {
      if (nxt[now][i] ^ -1) {
        v = nxt[now][i];
        q[r++] = v;
        f = fail[now];
        while (f && nxt[f][i] == -1) f = fail[f];
        fail[v] = (now && nxt[f][i] ^ -1) ? nxt[f][i] : 0;
      } else
        nxt[now][i] = (nxt[fail[now]][i] ^ -1) ? nxt[fail[now]][i] : 0;
    }
    f = fail[now];
    while (f && !tag[f]) f = ok[f];
    ok[now] = f;
  }
}
