"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>
from collections import defaultdict
letters = 'ACGT'
P = 10**9 + 9
n,m = map(int,raw_input().split())
v = [raw_input() for _ in xrange(m)]
d = {(0,''):0}
di = {0:(0,'')}
c = 1
a = defaultdict(dict)
t = set(v)
def f(k,s):
    global c
    if (k,s) not in d:
        d[k,s]=c
        di[c]=k,s
        c+=1
    return d[k,s]
def add(f,l,t):
    global a
    if l not in a[f] or di[a[f][l]][0]<di[t][0]:
        #print di[f], '->', di[t]
        a[f][l]=t
for s in v:
    for j in xrange(len(s)):
        for k in xrange(j+1):
            if j==len(s)-1: add(f(k,s[:j]),s[j],f(len(s),s[:j+1]))
            else: add(f(k,s[:j]),s[j],f(k,s[:j+1]))

for p,s in d.keys():
    for x in letters:
        z = s+x
        fm = False
        for j in v:
            if z.endswith(j) and p+len(j)>=len(z):
                fm = True
                break
        for j in xrange(len(z)):
            f = p-j,z[j:]
            if fm: f = len(z[j:]),z[j:]
            if f in d:
                add(d[p,s],x,d[f])
                break
#print d
#print a
d0,v0 = d,v
d = defaultdict(int)
d[0] = 1
for i in xrange(n):
    dn = defaultdict(int)
    for j in xrange(c):
        for x,v in a[j].iteritems():
            dn[v]+=d[j]
            dn[v]%=P
    #print ' '.join('%s=%d'%(di[k],v) for k,v in dn.iteritems() if v)
    d = dn

#print [di[k] for k,v in d.iteritems() if di[k][0]==len(di[k][1])]
print sum(v for k,v in d.iteritems() if di[k][0]==len(di[k][1]))%P