"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;
using ld = double;
using ll = long long;
using ull = unsigned long long;
const int INF = 1 << 30;
const ll LINF = 1LL << 60;
const int SZ = 1 << 13;
char buff[SZ], *pos = buff + SZ - 1;
inline ll read() {
  ll x = 0;
  int f = 1,
      c = (++pos == buff + SZ ? fread(pos = buff, 1, SZ, stdin), *pos : *pos);
  for (; !isdigit(c);
       c = (++pos == buff + SZ ? fread(pos = buff, 1, SZ, stdin), *pos : *pos))
    if (c == '-') f = -f;
  for (; isdigit(c);
       c = (++pos == buff + SZ ? fread(pos = buff, 1, SZ, stdin), *pos : *pos))
    x = (x << 3) + (x << 1) + (c ^ 48);
  return x * f;
}
const int MOD = 1e9 + 9;
const char *S = "ACGT";
int nxt[105][4];
int fail[105];
int len[105];
int cnt;
int n, m;
int dp[2][105][15];
void ins() {
  int u = 0, L = 0;
  for (int c =
           (++pos == buff + SZ ? fread(pos = buff, 1, SZ, stdin), *pos : *pos);
       !isspace(c); c = (++pos == buff + SZ ? fread(pos = buff, 1, SZ, stdin),
                         *pos               : *pos)) {
    ++L;
    int cc;
    for (int i = 0; i < 4; ++i)
      if (S[i] == c) {
        cc = i;
        break;
      }
    if (!nxt[u][cc]) nxt[u][cc] = ++cnt;
    u = nxt[u][cc];
  }
  len[u] = L;
}
void build() {
  queue<int> que;
  for (int i = 0; i < 4; ++i)
    if (nxt[0][i]) {
      fail[nxt[0][i]] = 0;
      que.push(nxt[0][i]);
    }
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (!len[u]) len[u] = len[fail[u]];
    for (int i = 0; i < 4; ++i)
      if (nxt[u][i]) {
        fail[nxt[u][i]] = nxt[fail[u]][i];
        que.push(nxt[u][i]);
      } else
        nxt[u][i] = nxt[fail[u]][i];
  }
}
void I(int p) { memset(dp[p], 0, sizeof dp[p]); }
void ADD(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int main() {
  n = read(), m = read();
  for (int i = 0; i < m; ++i) ins();
  build();
  int p = 0, q = 1;
  I(p);
  I(q);
  dp[p][0][1] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= cnt; ++j) {
      for (int k = 0; k < 15; ++k) {
        for (int c = 0; c < 4; ++c) {
          if (len[nxt[j][c]] >= k)
            ADD(dp[q][nxt[j][c]][1], dp[p][j][k]);
          else if (k + 1 < 15)
            ADD(dp[q][nxt[j][c]][k + 1], dp[p][j][k]);
        }
      }
    }
    swap(p, q);
    I(q);
  }
  int ans = 0;
  for (int i = 0; i <= cnt; ++i) ADD(ans, dp[p][i][1]);
  printf("%d\n", ans);
  return 0;
}
