[ABC246F] typewriter

Problem Statement

We have a typewriter with $N$ rows. The keys in the $i$-th row from the top can type the characters in a string $S_i$.

Let us use this keyboard to enter a string, as follows.

• First, choose an integer $1 \le k \le N$.
• Then, start with an empty string and only use the keys in the $k$-th row from the top to enter a string of length exactly $L$.

How many strings of length $L$ can be entered in this way? Since the answer can be enormous, print it modulo $998244353$.

Constraints

• $N$ and $L$ are integers.
• $1 \le N \le 18$
• $1 \le L \le 10^9$
• $S_i$ is a (not necessarily contiguous) non-empty subsequence of abcdefghijklmnopqrstuvwxyz.

---

Input

Input is given from Standard Input in the following format:

$N$ $L$
$S_1$
$S_2$
$\dots$
$S_N$

Output

Print the answer.

---

Sample Input 1

2 2
ab
ac

Sample Output 1

7

We can enter seven strings: aa, ab, ac, ba, bb, ca, cc.

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Sample Input 2

4 3
abcdefg
hijklmnop
qrstuv
wxyz

Sample Output 2

1352

---

Sample Input 3

5 1000000000
abc
acde
cefg
abcfh
dghi

Sample Output 3

346462871

Be sure to print the answer modulo $998244353$.

如果只有一个串，这一个串总共有 $x$ 个字符，那么构成的长度为 $l$ 的串总共有 $x^l$ 个。

然后我们很快发现会有重复的，那么就需要容斥原理。假设有 $k$ 个串，设在每个串中都出现了的字符数量为 $s$，那么他们共同重复的串为 $s^l$

然后容斥就可以了。

#include<bits/stdc++.h>
const int N=35,P=998244353;
char s[N];
int t[N],n,len,l;
long long ans;
int pow(int x,int y)
{
	if(!y)
		return 1;
	int t=pow(x,y>>1);
	if(y&1)
		return 1LL*t*t%P*x%P;
	return 1LL*t*t%P;
}
int mo(int x)
{
	return (x%P+P)%P;
}
int bitcnt(int x)
{
	int cnt=0;
	while(x)
	{
		x-=x&-x;
		++cnt;
	}
	return cnt;
}
void dfs(int x,int y,int z)
{
	if(x>n)
	{
		if(y)
		{
			int f=y&1? 1:-1;
			ans=mo(ans+f*pow(bitcnt(z),l)%P);
		}
	}
	else
	{
		dfs(x+1,y+1,z&t[x]);
		dfs(x+1,y,z);
	}
}
int main()
{
	scanf("%d%d",&n,&l);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s+1);
		len=strlen(s+1);
		for(int j=1;j<=len;j++)
			t[i]|=1<<s[j]-'a';
//		printf("%d\n",t[i]);
	}
	dfs(1,0,(1<<26)-1);
	printf("%lld",ans);
}