"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 mod = 1000000009;
struct node {
  int chi[4], fail, dep, edDep;
  node() {
    memset(chi, 0, sizeof(chi));
    dep = fail = edDep = 0;
  }
} a[105];
int V;
namespace acAuto {
void init() { V = 1; }
void ins(int* s, int n) {
  int u = 1;
  for (int i = 1; i <= n; i++) {
    if (!a[u].chi[s[i]]) {
      a[u].chi[s[i]] = ++V;
      a[V].dep = i;
    }
    u = a[u].chi[s[i]];
  }
  a[u].edDep = n;
}
void build() {
  queue<int> que;
  que.push(1);
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = 0; i < 4; i++) {
      int v = a[u].chi[i], f = a[u].fail;
      if (v) {
        if (f)
          a[v].fail = a[f].chi[i];
        else
          a[v].fail = 1;
        a[v].edDep = max(a[v].edDep, a[a[v].fail].edDep);
        que.push(v);
      } else {
        if (f)
          a[u].chi[i] = a[f].chi[i];
        else
          a[u].chi[i] = 1;
      }
    }
  }
}
}  // namespace acAuto
int idx[135], n, m, f[1005][105][15], sidx[15];
char s[15];
int main() {
  idx['A'] = 0, idx['C'] = 1, idx['G'] = 2, idx['T'] = 3;
  acAuto::init();
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s + 1);
    int l = strlen(s + 1);
    for (int j = 1; j <= l; j++) sidx[j] = idx[(int)s[j]];
    acAuto::ins(sidx, l);
  }
  acAuto::build();
  f[0][1][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int u = 1; u <= V; u++) {
      for (int j = 0; j <= a[u].dep; j++) {
        for (int k = 0; k < 4; k++) {
          int v = a[u].chi[k];
          if (a[v].dep >= j + 1) {
            (f[i + 1][v][a[v].edDep >= j + 1 ? 0 : j + 1] += f[i][u][j]) %= mod;
          }
        }
      }
    }
  }
  int ans = 0;
  for (int u = 1; u <= V; u++) (ans += f[n][u][0]) %= mod;
  printf("%d\n", ans);
  return 0;
}
