"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;
struct Vertex {
  Vertex *go[4];
  Vertex *pr, *suf;
  int came, num;
  int g;
  int end;
  Vertex() {
    memset(go, 0, sizeof(go));
    end = 0;
  }
};
const int maxn = 1000;
Vertex vs[maxn];
int vn;
const int maxs = 11;
char s[maxs];
void add() {
  Vertex *v = (&vs[0]);
  for (int i = 0; s[i]; i++) {
    int q = -1;
    switch (s[i]) {
      case 'A':
        q = 0;
        break;
      case 'C':
        q = 1;
        break;
      case 'G':
        q = 2;
        break;
      case 'T':
        q = 3;
        break;
      default:
        assert(0);
    }
    assert(q != -1);
    if (!((v->go)[q])) {
      vs[vn] = Vertex();
      (v->go)[q] = &vs[vn];
      vs[vn].pr = v;
      vs[vn].came = q;
      vs[vn].num = vn;
      vs[vn].end = 0;
      vs[vn].g = (v->g) + 1;
      vn++;
    }
    v = (v->go)[q];
  }
  v->end = (v->g);
}
Vertex *st[maxn];
const int MOD = 1000000009;
inline int mmod(int x) { return x >= MOD ? x - MOD : x; }
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) >= 2) {
    vn = 1;
    vs[0] = Vertex();
    vs[0].pr = vs[0].suf = (&vs[0]);
    vs[0].came = -1;
    vs[0].num = 0;
    vs[0].end = 0;
    vs[0].g = 0;
    for (int i = 0; i < m; i++) {
      scanf("%s", s);
      add();
    }
    assert(vn < maxn);
    st[0] = &vs[0];
    for (int l = 0, r = 1; l < r; l++) {
      Vertex *v = st[l]->pr;
      int came = st[l]->came;
      while (1) {
        if (v == (&vs[0])) {
          st[l]->suf = (&vs[0]);
          break;
        }
        assert(l);
        v = (v->suf);
        if ((v->go)[came]) {
          st[l]->suf = (v->go)[came];
          if (!(st[l]->end)) st[l]->end = (st[l]->suf)->end;
          break;
        }
      }
      for (int i = 0; i < 4; i++) {
        if ((st[l]->go)[i]) {
          st[r++] = (st[l]->go)[i];
        }
      }
    }
    vector<vector<int> > d(vn, vector<int>(maxs, 0));
    vector<vector<int> > dn = d;
    for (int i = 0; i < vn; i++) d[i][0] = 1;
    for (int iter = n - 1; iter >= 0; iter--) {
      for (int vnum = 0; vnum < vn; vnum++) {
        Vertex *v = &vs[vnum];
        for (int cov = 0; cov < maxs; cov++) {
          int resn = 0;
          for (int g = 0; g < 4; g++) {
            Vertex *vnow = v;
            bool ok = 1;
            while (!((vnow->go)[g])) {
              if (vnow == (&vs[0])) {
                ok = 0;
                break;
              }
              vnow = (vnow->suf);
            }
            if (!ok) continue;
            vnow = (vnow->go)[g];
            int newcov = cov + 1;
            if ((vnow->end) >= newcov) newcov = 0;
            if (newcov < maxs) resn = mmod(resn + d[vnow->num][newcov]);
          }
          dn[vnum][cov] = resn;
        }
      }
      swap(dn, d);
    }
    int ans = d[0][0];
    printf("%d\n", ans);
  }
  return 0;
}
