"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 maxN = 1011;
const int maxS = 110;
const int MOD = 1000000009;
int dp[maxN][12][maxS];
int pre[maxS], q[maxS], son[maxS][4], nxt[maxS][4], dang[maxS], len[maxS];
int n, m;
string st[11];
int tt;
inline int id(char ch) {
  if (ch == 'A') return 0;
  if (ch == 'C') return 1;
  if (ch == 'T') return 2;
  return 3;
}
void ins(string st) {
  int u = 1;
  for (int i = 0; i < st.size(); ++i) {
    if (son[u][id(st[i])] == 0) {
      ++tt;
      len[tt] = len[u] + 1;
      son[u][id(st[i])] = tt;
    }
    u = son[u][id(st[i])];
  }
  dang[u] = len[u];
}
int find(int u, int ch) {
  while (u > 1 && !son[u][ch]) u = pre[u];
  if (son[u][ch] > 0)
    return son[u][ch];
  else
    return u;
}
void bfs() {
  int he = 0, ta = 1;
  q[1] = 1;
  pre[1] = 1;
  while (he < ta) {
    he++;
    int u = q[he];
    for (int ch = 0; ch < 4; ++ch)
      if (son[u][ch]) {
        int v = son[u][ch];
        if (u == 1)
          pre[v] = 1;
        else
          pre[v] = find(pre[u], ch);
        if (dang[v] == 0) dang[v] = dang[pre[v]];
        q[++ta] = v;
      }
  }
}
void init() {
  cin >> n >> m;
  tt = 1;
  for (int i = 0; i < m; ++i) {
    cin >> st[i];
    ins(st[i]);
  }
  bfs();
  for (int i = 1; i <= tt; ++i)
    for (int j = 0; j < 4; ++j) {
      nxt[i][j] = find(i, j);
    }
}
inline void upd(int &x, int y) { x = (x + y) % MOD; }
void solve() {
  dp[0][0][1] = 1;
  for (int i = 0; i < n; ++i)
    for (int k = 0; k <= 10; ++k)
      for (int j = 1; j <= tt; ++j)
        if (dp[i][k][j] > 0) {
          for (int l = 0; l < 4; ++l) {
            int t = nxt[j][l];
            if (k + 1 - len[t] > 0) continue;
            if (dang[t] >= k + 1)
              upd(dp[i + k + 1][0][t], dp[i][k][j]);
            else
              upd(dp[i][k + 1][t], dp[i][k][j]);
          }
        }
  int ans = 0;
  for (int i = 1; i <= tt; ++i)
    if (dang[i]) upd(ans, dp[n][0][i]);
  printf("%d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}
