"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 = 1e6 + 11;
const double pi = acos(-1.0);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
struct Trie_tree {
  int fail;
  int son[26];
  int end;
  int mx;
} AC[111];
int siz = 0;
void insert(string s, int val) {
  int len = s.size();
  int u = 0;
  for (int(i) = (0); (i) < (len); (i)++) {
    if (!AC[u].son[s[i] - 'A']) AC[u].son[s[i] - 'A'] = ++siz;
    u = AC[u].son[s[i] - 'A'];
  }
  AC[u].end++;
  AC[u].mx = val;
}
void build_fail() {
  queue<int> Q;
  while (!Q.empty()) Q.pop();
  for (int(i) = (0); (i) < (26); (i)++)
    if (AC[0].son[i]) {
      Q.push(AC[0].son[i]);
      AC[AC[0].son[i]].fail = 0;
    }
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int(i) = (0); (i) < (26); (i)++)
      if (AC[u].son[i]) {
        AC[AC[u].son[i]].fail = AC[AC[u].fail].son[i];
        AC[AC[u].son[i]].mx =
            max(AC[AC[u].son[i]].mx, AC[AC[AC[u].fail].son[i]].mx);
        Q.push(AC[u].son[i]);
      } else
        AC[u].son[i] = AC[AC[u].fail].son[i];
  }
}
int n, m;
int dp[1001][1001][11];
bool vis[1001][1001][11];
char ch[4] = {'A', 'C', 'G', 'T'};
int main() {
  scanf("%d%d", &n, &m);
  for (int(i) = (0); (i) < (m); (i)++) {
    string s;
    cin >> s;
    insert(s, s.size());
  }
  build_fail();
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  vis[0][0][0] = 1;
  queue<pair<int, pair<int, int> > > Q;
  Q.push(make_pair(0, make_pair(0, 0)));
  while (!Q.empty()) {
    int i = Q.front().first;
    int j = Q.front().second.first;
    int k = Q.front().second.second;
    Q.pop();
    if (i == n) continue;
    for (int(l) = (0); (l) < (4); (l)++) {
      int u = AC[j].son[ch[l] - 'A'];
      if (AC[u].mx > k) {
        dp[i + 1][u][0] = (dp[i + 1][u][0] + dp[i][j][k]) % 1000000009;
        if (!vis[i + 1][u][0]) {
          Q.push(make_pair(i + 1, make_pair(u, 0)));
          vis[i + 1][u][0] = 1;
        }
      } else {
        if (k >= 9) continue;
        dp[i + 1][u][k + 1] = (dp[i + 1][u][k + 1] + dp[i][j][k]) % 1000000009;
        if (!vis[i + 1][u][k + 1]) {
          Q.push(make_pair(i + 1, make_pair(u, k + 1)));
          vis[i + 1][u][k + 1] = 1;
        }
      }
    }
  }
  int ans = 0;
  for (int(i) = (0); (i) < (siz + 1); (i)++)
    ans = (ans + dp[n][i][0]) % 1000000009;
  printf("%d\n", ans);
  return 0;
}
