"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 N = 15;
const int M = 1050;
const int MOD = 1000000009;
int n, m;
char mp[15];
int ci(char ch) {
  if (ch == 'A') return 0;
  if (ch == 'T') return 1;
  if (ch == 'G') return 2;
  return 3;
}
int rt[N];
int dp[2][105][M];
void Mod(int &x) {
  if (x >= MOD) x -= MOD;
}
struct Trie {
  int tot, ch[105][4], fl[105], ln[105], cl[105], hed[105], cnt;
  struct EG {
    int to, nxt;
  } e[105];
  void ae(int f, int t) {
    e[++cnt].to = t;
    e[cnt].nxt = hed[f];
    hed[f] = cnt;
  }
  int insert() {
    int len = strlen(mp + 1);
    int u = 0;
    for (int i = 1; i <= len; i++) {
      int c = ci(mp[i]);
      if (!ch[u][c]) ch[u][c] = ++tot, ln[tot] = ln[u] + 1;
      u = ch[u][c];
    }
    return u;
  }
  void build() {
    queue<int> q;
    for (int i = 0; i < 4; i++)
      if (ch[0][i]) q.push(ch[0][i]), ae(0, ch[0][i]);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int i = 0; i < 4; i++) {
        int &v = ch[u][i];
        if (!v) {
          v = ch[fl[u]][i];
          continue;
        }
        fl[v] = ch[fl[u]][i];
        ae(fl[v], v);
        q.push(v);
      }
    }
  }
  void init() {
    for (int i = 1; i <= m; i++) cl[rt[i]] = ln[rt[i]];
    queue<int> q;
    q.push(0);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int j = hed[u]; j; j = e[j].nxt) {
        int to = e[j].to;
        cl[to] = max(cl[to], cl[u]);
        q.push(to);
      }
    }
  }
  int DP() {
    dp[0][0][0] = 1;
    for (int I = 0, i = 0; I < n; memset(dp[i], 0, sizeof(dp[i])), I++, i ^= 1)
      for (int j = 0; j <= tot; j++)
        for (int k = 0; k <= I; k++)
          if (dp[i][j][k])
            for (int c = 0; c < 4; c++)
              if (ch[j][c]) {
                if (cl[ch[j][c]] >= k + 1)
                  Mod(dp[i ^ 1][ch[j][c]][0] += dp[i][j][k]);
                else
                  Mod(dp[i ^ 1][ch[j][c]][k + 1] += dp[i][j][k]);
              }
    int ans = 0;
    for (int i = 0; i <= tot; i++) Mod(ans += dp[n & 1][i][0]);
    return ans;
  }
} tr;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", mp + 1);
    rt[i] = tr.insert();
  }
  tr.build();
  tr.init();
  int ans = tr.DP();
  printf("%d\n", ans);
  return 0;
}
