直接容斥即可，每次选出它们的并集。
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#define fo(i,a,b) for (int (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (int (i)=(b);(i)>=(a);(i)--)
#define mk(x,y) make_pair((x),(y))
using namespace std;
typedef double db;
typedef long long ll;
const int N=105;
const ll mo=998244353;
const ll inf=1ll<<60;
ll b[N],s[N],l,n,len,ss,tot,z,y,f[N],ans;
char t[N];
ll power(ll a,ll b){
ll t=1,y=a%mo;
while (b){
if (b&1) t=t*y%mo;
y=y*y%mo;
b/=2;
}
return t;
}
int main()
{
// freopen("data.in","r",stdin);
b[0]=1;
fo(i,1,26) b[i]=b[i-1]*2;
scanf("%lld %lld",&n,&l);
fo(i,1,26) f[i]=power(i,l);
fo(p,0,n-1) {
scanf("%s",t+1);
len=strlen(t+1);
fo(i,1,len) {
s[p]|=b[t[i]-'a'];
}
}
fo(st,1,(1<<n)-1) {
tot=0;
ss=b[26]-1;
fo(i,0,n-1) if (b[i]&st) ss&=s[i],tot++;
y=0;
fo(i,0,25) if (ss&b[i]) y++;
if (tot & 1) ans=(ans+f[y])%mo;
else ans=(ans-f[y])%mo;
}
ans=(ans%mo+mo)%mo;
printf("%lld",ans);
return 0;
}