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 templateusing pii = pair<int, int>;#define fir first#define sec secondTp<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_backTp<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 longint 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.}