Problem: A chainword is a special type of crossword. As most of the crosswords do, it has cells that you put the letters in and some sort of hints to what these letters should be.

The letter cells in a chainword are put in a single row. We will consider chainwords of length m in this task.

A hint to a chainword is a sequence of segments such that the segments don't intersect with each other and cover all m letter cells. Each segment contains a description of the word in the corresponding cells.

The twist is that there are actually two hints: one sequence is the row above the letter cells and the other sequence is the row below the letter cells. When the sequences are different, they provide a way to resolve the ambiguity in the answers.

You are provided with a dictionary of n words, each word consists of lowercase Latin letters. All words are pairwise distinct.

An instance of a chainword is the following triple: 

  * a string of m lowercase Latin letters; 
  * the first hint: a sequence of segments such that the letters that correspond to each segment spell a word from the dictionary; 
  * the second hint: another sequence of segments such that the letters that correspond to each segment spell a word from the dictionary. 



Note that the sequences of segments don't necessarily have to be distinct.

Two instances of chainwords are considered different if they have different strings, different first hints or different second hints.

Count the number of different instances of chainwords. Since the number might be pretty large, output it modulo 998 244 353.

Input

The first line contains two integers n and m (1 ≤ n ≤ 8, 1 ≤ m ≤ 10^9) — the number of words in the dictionary and the number of letter cells.

Each of the next n lines contains a word — a non-empty string of no more than 5 lowercase Latin letters. All words are pairwise distinct. 

Output

Print a single integer — the number of different instances of chainwords of length m for the given dictionary modulo 998 244 353.

Examples

Input


3 5
ababa
ab
a


Output


11


Input


2 4
ab
cd


Output


4


Input


5 100
a
aa
aaa
aaaa
aaaaa


Output


142528942

Note

Here are all the instances of the valid chainwords for the first example: 

<image>

The red lines above the letters denote the segments of the first hint, the blue lines below the letters denote the segments of the second hint.

In the second example the possible strings are: "abab", "abcd", "cdab" and "cdcd". All the hints are segments that cover the first two letters and the last two letters.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 998244353;

template<ll P>
struct ModNum {
	ll v;
	ModNum(ll val) : v(val) {}
	operator ll() const { return v; }

	ModNum operator+(const ModNum& rhs) { ll res = v + rhs.v; return res >= MOD ? res - MOD : res; }
	ModNum operator-(const ModNum& rhs) { ll res = v - rhs.v; return res >= 0 ? res : res + MOD; }
	ModNum operator*(const ModNum& rhs) { return v * rhs.v % MOD; }
	ModNum operator/(const ModNum& rhs) { return v * rhs.pow(MOD - 2) % MOD; }
	ModNum& operator+=(const ModNum& rhs) { v += rhs.v; if (v >= MOD) v -= MOD; return *this; }
	ModNum& operator-=(const ModNum& rhs) { v -= rhs.v; if (v < 0) v += MOD; return *this; }
	ModNum& operator*=(const ModNum& rhs) { v = v * rhs.v % MOD; return *this; }
	ModNum& operator/=(const ModNum& rhs) { v = v * rhs.pow(MOD - 2) % MOD; return *this; }
	
	ModNum pow(ll b) const {
		ModNum res(1), cur(v);
		for (ll j = 1; j <= b; j <<= 1) {
			if (b & j) res *= cur;
			cur *= cur;
		}
		return res;
	}
};
using num = ModNum<MOD>;

constexpr int bLog(ll v) { return v ? 63 - __builtin_clzll(v) : 0; }
template<class T>
T itPow(const T& a, ll d, T r = 1) {
        for (ll b = 1ll << bLog(d); b > 0; b /= 2) {
                r = r*r;
                if (d & b) r = r*a;
        }
        return r;
}

// Assumes 0 and 1 are 0,1 for T
template<class T>
struct Matrix {
	vector<T> mat;
	int h, w;

	Matrix(int height, int width) : h(height), w(width), mat(height*width, 0) {}

	const T& operator()(int y, int x) const {
		return mat[x+y*w];
	}
	T& operator()(int y, int x) {
		return mat[x+y*w];
	}
	static Matrix ident(int n) {
		Matrix res(n, n);
		for (int i = 0; i < n; ++i) res(i, i) = 1;
		return res;
	}
	Matrix& operator+=(const Matrix& rhs) {
		assert(h == rhs.h && w == rhs.w);
		for (int i = 0; i < w*h; ++i) mat[i] += rhs.mat[i];
	}
	Matrix operator*(const Matrix& rhs) const {
		assert(w == rhs.h);
		Matrix res(h, rhs.w);
		for (int i = 0; i < w; ++i) {
			int b1 = i * rhs.w;
			for (int y = 0; y < res.h; ++y) {
				T mult = mat[i + y * w];
				int b2 = y * res.w;
				for (int x = 0; x < res.w; ++x) {
					res.mat[b2 + x] += mult * rhs.mat[b1 + x];
				}
			}
		}
		return res;
	}
};
using Mat = Matrix<num>;

// returns number of elements strictly smaller than v in vec
template<class T>
int bins(const vector<T>& vec, T v) {
	int low = 0;
	int high = vec.size();
	while(low != high) {
		int mid = (low + high) >> 1;
		if (vec[mid] < v) low = mid + 1;
		else high = mid;
	}
	return low;
}

string getSuff(const string& str, int k) {
	string res(k, '_');
	for (int i = 0; i < k; ++i) res[i] = str[(int)str.size() - k + i];
	return res;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	// Over every string of length m, add (#ways to make it by appending words from dictionary)^2

	// Current state:
	//	Current number of characters
	//	suffix of some word that is currently unfilled
	// 	

	int n, m;
	cin >> n >> m;

	vector<string> words(n);
	for (int i = 0; i < n; ++i) cin >> words[i];
	
	vector<string> states;
	for (int i = 0; i < n; ++i) {
		for (int k = 1; k <= words[i].size(); ++k) {
			states.emplace_back(getSuff(words[i], k));
		}
	}
	states.emplace_back("");

	sort(states.begin(), states.end());
	states.erase(unique(states.begin(), states.end()), states.end());

	int ss = states.size();
	
	/*
	cerr << ss << ": ";
	for (auto str : states) cerr << "[" << str << "] "; cerr << '\n';
	*/

	vector<pair<int, int>> ord(ss);
	for (int i = 0; i < ss; ++i) ord[i] = {(int)states[i].size(), i};
	sort(ord.begin(), ord.end());

	vector<num> zw(ss, 0);
	zw[0] = 1;

	vector<vector<pair<int, int>>> ts(ss);
	for (auto pr : ord) {
		int i = pr.second;
		string stat = states[i];
		for (int j = 0; j < words.size(); ++j) {
			string wrd = words[j];
			if (wrd.empty()) continue;

			bool works = 1;
			for (int t = 0; t < min(stat.size(), wrd.size()); ++t) {
				if (stat[t] != wrd[t]) works = 0;
			}
			if (works) {
				if (stat.size() >= wrd.size()) {
					string rem = getSuff(stat, (int)stat.size() - (int)wrd.size());
					int t = bins(states, rem);
					zw[i] += zw[t];
					for (auto pr2 : ts[t]) {
						ts[i].emplace_back(pr2.first, pr2.second);
					}
				} else {
					string rem = getSuff(wrd, (int)wrd.size() - (int)stat.size());
					int t = bins(states, rem);
					ts[i].emplace_back(t, (int)rem.size());
				}
			}
		}
		// cerr << i << ' ' << zw[i] << '\n';
	}

	/*
	for (int s = 0; s < ss; ++s) {
		cerr << states[s] << " " << zw[s] << ": ";
		for (auto pr : ts[s]) cerr << states[pr.first] << ' ' << pr.second << ", ";
		cerr << '\n';
	}
	*/

	Mat mat(5 * ss, 5 * ss);
	for (int y = 0; y < ss; ++y) {
		for (int j = 0; j < 4; ++j) {
			int b = y + j * ss;
			mat(b + ss, b) = 1;
		}
	}
	for (int s = 0; s < ss; ++s) {
		for (auto pr : ts[s]) {
			mat((5 - pr.second)*ss + s, 4*ss + pr.first) += 1;
		}
	}
	
	/*
	for (int y = 0; y < 5*ss; ++y) {
		for (int x = 0; x < 5*ss; ++x) {
			cerr << mat(y, x) << ' ';
		}
		cerr << '\n';
	}
	*/

	Mat res = itPow(mat, m, mat.ident(5 * ss));

	/*
	for (int y = 0; y < 5*ss; ++y) {
		for (int x = 0; x < 5*ss; ++x) {
			cerr << res(y, x) << ' ';
		}
		cerr << '\n';
	}
	*/

	num ans = 0;
	for (int s = 0; s < ss; ++s) ans += res(4*ss, 4*ss + s) * zw[s];
	cout << ans << '\n';
}