"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);
int n, m;
int pos;
char seq[11];
int des[101][26], nxt[101], q[100], depth[101], lEnd[101], dp[1001][101][10];
int Insert(const char* const p) {
  int l, now(0);
  for (l = 0; p[l]; ++l)
    if (des[now][p[l] - 'A'])
      now = des[now][p[l] - 'A'];
    else {
      des[now][p[l] - 'A'] = ++pos;
      depth[pos] = l + 1;
      now = pos;
    }
  lEnd[now] = l;
  return l;
}
void Build(void) {
  int l(0), r(0), now;
  for (int i = 0; i < 26; ++i) des[0][i] && (q[r++] = des[0][i]);
  while (l < r) {
    now = q[l];
    for (int i = 0; i < 26; ++i) {
      int& t(des[now][i]);
      if (t) {
        nxt[t] = des[nxt[now]][i];
        q[r++] = t;
      } else
        t = des[nxt[now]][i];
    }
    ++l;
  }
  for (int i = 0; i < r; ++i) {
    int t(q[i]);
    lEnd[t] < lEnd[nxt[t]] && (lEnd[t] = lEnd[nxt[t]]);
  }
  return;
}
int main() {
  int maxl(0), t1, t2, t3, ans(0);
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    scanf("%s", seq);
    t1 = Insert(seq);
    maxl < t1 && (maxl = t1);
  }
  Build();
  dp[0][0][0] = 1;
  for (int l = 1; l <= n; ++l) {
    for (int stat = 0; stat <= pos; ++stat) {
      for (int rem = 0; rem < maxl; ++rem) {
        if (!dp[l - 1][stat][rem]) continue;
        for (int c = 0; c < 26; ++c) {
          t1 = des[stat][c];
          if (!t1) continue;
          t2 = depth[t1];
          if (!lEnd[stat] && t2 < rem + 1) continue;
          t3 = lEnd[t1] >= rem + 1 ? 0 : rem + 1;
          if (t3 == maxl) continue;
          dp[l][t1][t3] += dp[l - 1][stat][rem];
          dp[l][t1][t3] >= MOD && (dp[l][t1][t3] -= MOD);
        }
      }
    }
  }
  for (int i = 0; i <= pos; ++i)
    if (lEnd[i]) {
      ans += dp[n][i][0];
      ans >= MOD && (ans -= MOD);
    }
  printf("%d\n", ans);
  return 0;
}
