[JSOI2007]文本生成器 [AC自动机，dp]

时刻要记住正难则反，可以知道总数是 \(26^m\)，我们可以减掉不合法的。
AC自动机上面dp，不合法的显然就是没有出现任意的一个串，根据rainy的教导
单词 \(b,bce,abcd\) 的 ACAM

然后 \(dp\) 就好了，由于点数不超过 \(n*m \leq 6000\)，然后你每一位枚举复杂度是 \(m^2n\) 的
可以通过本题

// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define Tp template
using pii = pair<int, int>;
#define fir first
#define sec second
Tp<class T> void cmax(T& x, const T& y) {if (x < y) x = y;} Tp<class T> void cmin(T& x, const T& y) {if (x > y) x = y;}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T> void sort(vector<T>& v) { sort(all(v)); } Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
const int SZ = 1 << 23 | 233;
struct FILEIN { char qwq[SZ], *S = qwq, *T = qwq, ch;
#ifdef __WIN64
#define GETC getchar
#else
  char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
#endif
  FILEIN& operator>>(char& c) {while (isspace(c = GETC()));return *this;}
  FILEIN& operator>>(string& s) {while (isspace(ch = GETC())); s = ch;while (!isspace(ch = GETC())) s += ch;return *this;}
  Tp<class T> void read(T& x) { bool sign = 0;while ((ch = GETC()) < 48) sign ^= (ch == 45); x = (ch ^ 48);
    while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48); x = sign ? -x : x;
  }FILEIN& operator>>(int& x) { return read(x), *this; } FILEIN& operator>>(ll& x) { return read(x), *this; }
} in;
struct FILEOUT {const static int LIMIT = 1 << 22 ;char quq[SZ], ST[233];int sz, O;
  ~FILEOUT() { flush() ; }void flush() {fwrite(quq, 1, O, stdout); fflush(stdout);O = 0;}
  FILEOUT& operator<<(char c) {return quq[O++] = c, *this;}
  FILEOUT& operator<<(string str) {if (O > LIMIT) flush();for (char c : str) quq[O++] = c;return *this;}
  Tp<class T> void write(T x) {if (O > LIMIT) flush();if (x < 0) {quq[O++] = 45;x = -x;}
		do {ST[++sz] = x % 10 ^ 48;x /= 10;} while (x);while (sz) quq[O++] = ST[sz--];
  }FILEOUT& operator<<(int x) { return write(x), *this; } FILEOUT& operator<<(ll x) { return write(x), *this; }
} out;
#define int long long

int n , m ;

const int maxn = 66 ;
const int maxm = 111 ;

int dp[maxm][maxn * maxm] ;
const int mod = 10007 ;

int qpow(int x , int y) {
	int ans = 1 ;
	for( ; y ; y >>= 1 , x = x * x % mod)
		if(y & 1)
			ans = ans * x % mod ;
	return ans ; 
}

struct ACAM {
	int ch[maxn * maxm][26] , fail[maxn * maxm] , ed[maxn * maxm] , cnt = 1 ;
	
	void ins(string s) {
		int p = 1 ;
		for(char x : s) {
			int c = x - 'A' ;
			if(! ch[p][c]) ch[p][c] = ++ cnt ;
			p = ch[p][c] ;
		}
		ed[p] |= 1 ;
	}
	
	void build() {
		queue < int > q ;
		for(int i = 0 ; i < 26 ; i ++)
			if(ch[1][i])
				fail[ch[1][i]] = 1 , q.push(ch[1][i]) ;
			else
				ch[1][i] = 1 ;
		
		while(! q.empty()) {
			int u = q.front() ;
			q.pop() ;
			for(int i = 0 ; i < 26 ; i ++)
				if(ch[u][i])
					fail[ch[u][i]] = ch[fail[u]][i] , ed[ch[u][i]] |= ed[fail[ch[u][i]]] , q.push(ch[u][i]) ;
				else
					ch[u][i] = ch[fail[u]][i] ;
		}
	}
	
	int solve() {
		memset(dp , 0 , sizeof(dp)) ;
		dp[0][1] = 1 ;
		rep(i , 0 , m - 1)
			rep(j , 1 , cnt)
				rep(k , 0 , 25)
					if(! ed[ch[j][k]]) 
						(dp[i + 1][ch[j][k]] += dp[i][j]) %= mod ;
		int qwq = 0 ;
		rep(i , 1 , cnt) (qwq += dp[m][i]) %= mod ;
		return qwq ;
	}
} acam ;


signed main() {
  // code begin.
	in >> n >> m ;
	rep(i , 1 , n) {
		string s ;
		in >> s ;
		acam.ins(s) ;
	}
	acam.build() ;
	int ans = qpow(26 , m) ;
	(ans += mod - acam.solve()) %= mod ;
	out << ans << '\n' ;
	return 0;
  // code end.
}