"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 N = 1e3 + 5;
const int CNTV = 105;
const int LENSUFF = 11;
const int K = 4;
const int MOD = 1e9 + 9;
struct node {
  int go[K], suff, p, pch, lev, cr;
  node() {
    memset(go, -1, sizeof go);
    suff = p = pch = cr = -1;
    lev = 0;
  }
};
int sz = 1, num[256];
node a[CNTV];
inline void add(string second) {
  int v = 0;
  for (int i = 0; i < second.size(); ++i) {
    int c = num[second[i]];
    if (a[v].go[c] == -1) {
      a[sz].lev = a[v].lev + 1;
      a[sz].p = v;
      a[sz].pch = c;
      a[v].go[c] = sz++;
    }
    v = a[v].go[c];
  }
  a[v].cr = v;
}
int go(int v, int c);
int sufflink(int v);
int go(int v, int c) {
  if (a[v].go[c] != -1) return a[v].go[c];
  if (v == 0) return a[v].go[c] = 0;
  return a[v].go[c] = go(sufflink(v), c);
}
int sufflink(int v) {
  if (v == 0 || a[v].p == 0) return a[v].suff = 0;
  if (a[v].suff != -1) return a[v].suff;
  return a[v].suff = go(sufflink(a[v].p), a[v].pch);
}
int cross(int v) {
  if (a[v].cr != -1) return a[v].cr;
  return a[v].cr = cross(sufflink(v));
}
int n, m, dp[N][CNTV][LENSUFF];
int main() {
  num['A'] = 0;
  num['C'] = 1;
  num['G'] = 2;
  num['T'] = 3;
  a[0].cr = 0;
  cin >> n >> m;
  string second;
  for (int i = 0; i < m; ++i) {
    cin >> second;
    add(second);
  }
  dp[0][0][0] = 1;
  for (int len = 0; len < n; ++len)
    for (int v = 0; v < sz; ++v)
      for (int need = 0; need < 10; ++need)
        for (int c = 0; c < 4; ++c) {
          int to = go(v, c);
          if (a[to].lev >= need + 1) {
            int cr = cross(to);
            if (a[cr].lev >= need + 1)
              dp[len + 1][to][0] =
                  (dp[len + 1][to][0] + dp[len][v][need]) % MOD;
            else
              dp[len + 1][to][need + 1] =
                  (dp[len + 1][to][need + 1] + dp[len][v][need]) % MOD;
          }
        }
  int ans = 0;
  for (int v = 0; v < sz; ++v) ans = (ans + dp[n][v][0]) % MOD;
  cout << ans;
  return 0;
}
