"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;
using ll = long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
const int INF = 0x3f3f3f3f, N = 1e3 + 5;
const ll LINF = 1e18 + 5;
constexpr int mod = 1e9 + 9;
int n, m;
int id[150];
ll dp[2][105][N];
char s[N];
struct ACA {
  static const int M = 105;
  int nxt[M][26], fail[M], cnt, len[M], end[M];
  void insert(char *s, int idx) {
    int p = 0;
    for (int i = 0; s[i]; i++) {
      int k = id[s[i]];
      if (!nxt[p][k]) nxt[p][k] = ++cnt;
      p = nxt[p][k];
    }
    end[p] = 1;
    len[p] = strlen(s);
  }
  void build() {
    queue<int> q;
    for (int i = 0; i < 4; i++)
      if (nxt[0][i]) q.push(nxt[0][i]);
    while (!q.empty()) {
      int k = q.front();
      q.pop();
      for (int i = 0; i < 4; i++) {
        if (nxt[k][i]) {
          fail[nxt[k][i]] = nxt[fail[k]][i];
          if (end[fail[nxt[k][i]]])
            end[nxt[k][i]] = 1,
            len[nxt[k][i]] = max(len[nxt[k][i]], len[fail[nxt[k][i]]]);
          q.push(nxt[k][i]);
        } else
          nxt[k][i] = nxt[fail[k]][i];
      }
    }
  }
  void work() {
    dp[0][0][0] = 1;
    int c = 1;
    for (int i = 1; i <= n; i++, c ^= 1) {
      for (int j = 0; j <= cnt; j++)
        for (int k = max(0, i - 10); k <= i; k++) dp[c][j][k] = 0;
      for (int j = 0; j <= cnt; j++)
        for (int d = 0; d < 4; d++) {
          int nxtj = nxt[j][d];
          for (int k = max(0, i - 10); k < i; k++) {
            if (!end[nxtj])
              (dp[c][nxtj][k] += dp[c ^ 1][j][k]) %= mod;
            else {
              if (k + len[nxtj] < i)
                (dp[c][nxtj][k] += dp[c ^ 1][j][k]) %= mod;
              else
                (dp[c][nxtj][i] += dp[c ^ 1][j][k]) %= mod;
            }
          }
        }
    }
    ll ans = 0;
    for (int i = 0; i <= cnt; i++) ans = (ans + dp[c ^ 1][i][n]) % mod;
    if (ans < 0) ans += mod;
    printf("%lld\n", ans);
  }
} ac;
int main() {
  id['A'] = 0;
  id['C'] = 1;
  id['G'] = 2;
  id['T'] = 3;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s);
    ac.insert(s, i);
  }
  ac.build();
  ac.work();
  return 0;
}
