題目大意：
給定m個01串，問有多少長度為2n的反迴文01串，包含這m個串。
一個01串被稱為反迴文的，當且僅當 $\mathrm{\forall }i\in [1,n]$

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define Rep(i,v) rep(i,0,(int)v.size()-1)
#define lint long long
#define mod 998244353
#define ull unsigned lint
#define db long double
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define gc getchar()
#define debug(x) cerr<<#x<<"="<<x
#define sp <<" "
#define ln <<endl
using namespace std;
typedef pair<int,int> pii;
typedef set<int>::iterator sit;
inline int inn()
{
    int x,ch;while((ch=gc)<'0'||ch>'9');
    x=ch^'0';while((ch=gc)>='0'&&ch<='9')
        x=(x<<1)+(x<<3)+(ch^'0');return x;
}
const int LEN=110,TR=1210,MXS=(1<<6)+10,N=510;
char str[LEN];queue<int> q;int node_cnt,fa[TR],s[TR];
int a[LEN],fr[TR],ch[TR][2],frv[TR],dp[N][MXS][TR];
inline int ins(int *a,int n,int v)
{
    int x=1;
    for(int i=1,c;i<=n;fr[ch[x][c]]=x,x=ch[x][c],i++)
        if(!ch[x][c=a[i]]) ch[x][c]=++node_cnt,frv[ch[x][c]]=c;
    return s[x]|=v;
}
inline int getfail()
{
    while(!q.empty()) q.pop();
    rep(i,0,1)
    {
        int &c=ch[1][i];
        if(c) fa[c]=1,q.push(c);
        else c=1;
    }
    while(!q.empty())
    {
        int x=q.front();
        s[x]|=s[fa[x]],q.pop();
        rep(i,0,1)
        {
            int &y=ch[x][i],f=fa[x],c=ch[f][i];
            if(y) fa[y]=c,q.push(y);else y=c;
        }
    }
    return 0;
}
inline int gett(int x,int ans=0)
{
    for(int y=x;fr[x];x=fr[x]) ans|=s[y=ch[y][frv[x]^1]];
    return ans;
}
inline int upd(int &x,int y) { return x+=y,(x>=mod?x-=mod:0); }
int main()
{
    int m=inn(),n=inn();node_cnt=1;
    rep(qwq,1,m)
    {
        scanf("%s",str+1);int l=(int)strlen(str+1);
        rep(i,1,l) a[i]=str[i]-'0';ins(a,l,1<<(qwq-1));
        rep(i,1,l) a[i]=!a[i];
        rep(i,1,l/2) swap(a[i],a[l-i+1]);
        ins(a,l,1<<(qwq-1));
    }
    getfail(),dp[0][0][1]=1;
    int nc=node_cnt,all=(1<<m)-1;
    rep(i,0,n-1) rep(j,0,all) rep(k,1,nc)
        if(dp[i][j][k]) for(int t=0,p;t<=1;t++)
            p=ch[k][t],upd(dp[i+1][j|s[p]][p],dp[i][j][k]);
    int ans=0;
    rep(i,1,nc)
    {
        int t=gett(i);
        rep(j,0,all) if((j|t)==all) upd(ans,dp[n][j][i]);
    }
    return !printf("%d\n",ans);
}