"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>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
void read(T& x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
struct AC_Auto {
  static const int N = 110;
  static const long long p = 1000000009;
  int sz = 0;
  int tri[N][4], tri2[N][4], end[N], fail[N], last[N], last2[N];
  int dep[N];
  long long f[1010][11][N];
  int init(int id = 0) {
    memset(tri[id], 0, sizeof(tri[id]));
    memset(tri2[id], 0, sizeof(tri2[id]));
    end[id] = fail[id] = last[id] = last2[id] = dep[id] = 0;
    return sz = id;
  }
  int insert(const char* s) {
    int x = 0;
    for (int i = 0; s[i]; ++i) {
      int& v = tri[x][s[i] - 'A'];
      if (v == 0) {
        v = init(sz + 1);
        dep[v] = dep[x] + 1;
      }
      x = v;
    }
    end[x] = 1;
    return x;
  }
  void build() {
    for (int i = 0; i <= sz; ++i) memcpy(tri2[i], tri[i], sizeof(tri[i]));
    queue<int> q;
    for (auto v : tri[0]) {
      if (v) {
        fail[v] = 0;
        q.push(v);
      }
    }
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      if (end[x]) last2[x] = x;
      for (int i = 0; i < 4; ++i) {
        int& v = tri[x][i];
        if (v) {
          fail[v] = tri[fail[x]][i];
          last[v] = end[fail[v]] ? fail[v] : last[fail[v]];
          last2[v] = last2[x];
          q.push(v);
        } else {
          v = tri[fail[x]][i];
        }
      }
    }
  }
  void fuck(int n) {
    vector<int> nodes;
    for (int i = 0; i <= sz; ++i) nodes.push_back(i);
    sort(nodes.begin(), nodes.end(),
         [&](int a, int b) { return dep[a] > dep[b]; });
    f[0][0][0] = 1;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 10; ++j) {
        for (auto v : nodes) {
          long long tmp = f[i][j][v];
          if (tmp == 0) continue;
          for (int c = 0; c < 4; ++c) {
            int u = v;
            int nj = j;
            while (true) {
              int x = u;
              u = tri2[x][c];
              if (u) {
                ++nj;
                if (end[u] || dep[last[u]] >= nj) nj = 0;
                f[i + 1][nj][u] = (f[i + 1][nj][u] + tmp) % p;
                break;
              }
              if (x == 0) break;
              u = fail[x];
              if (dep[u] >= nj)
                nj = min(nj, dep[u] - dep[last2[u]]);
              else
                break;
            }
          }
        }
      }
    }
    long long ans = 0;
    for (auto v : nodes) {
      ans += f[n][0][v];
      for (int i = 1; i < 10; ++i) {
        int x = fail[v];
        if (dep[x] >= i) {
          int ni = min(i, dep[x] - dep[last2[x]]);
          f[n][ni][x] += f[n][i][v];
        }
      }
    }
    cout << ans % p;
  }
} AC;
char s[20];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  while (m--) {
    cin >> s;
    for (int i = 0; s[i]; ++i) {
      if (s[i] == 'G')
        s[i] = 'B';
      else if (s[i] == 'T')
        s[i] = 'D';
    }
    AC.insert(s);
  }
  AC.build();
  AC.fuck(n);
  return 0;
}
