// Marcin Knapik
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;

#define sz(x) ((int)(x).size())
#define pb push_back
#define all(s) s.begin(), s.end()
#define f first
#define s second

const int N = 1e6 + 6;
const int mod = 998244353;

#define vi vector<int>
#define vvi vector<vi>

int f(int a, int b, int c){
    return a * 40 + b * 8 + (a == b ? 0 : c);
}

vvi operator * (vvi & a, vvi & b){
    int n = sz(a);
    assert(sz(a) and sz(b));
    assert(sz(a) == sz(a[0]) and sz(a[0]) == sz(b) and sz(b) == sz(b[0]));
    vvi ret(n, vi(n, 0));
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            for(int k = 0; k < n; k++)
                ret[i][j] = (ret[i][j] + 1ll * a[i][k] * b[k][j]) % mod;
    return ret;
}

vvi fast(vvi a, int b){
    vvi ret = a;
    for(int i = 0; i < sz(ret); i++)
        for(int j = 0; j < sz(ret); j++)
            ret[i][j] = (i == j);

    while(b){
        if(b & 1)
            ret = ret * a;
        a = a * a;
        b /= 2;
    }

    return ret;
}

void solve(){
    int n, m;
    cin >> n >> m;
    vector<string> tab(n);

    for(int i = 0; i < n; i++)
        cin >> tab[i];

    vvi row(5 * 5 * 8, vi(5 * 5 * 8));
    row[0][f(0, 0, 0)] = 1;

    vvi mat(5 * 5 * 8, vi(5 * 5 * 8));

    for(int i = 1; i <= 4; i++)
        for(int j = 1; j <= 4; j++)
            for(int k = 0; k < n; k++)
                mat[f(i, j, k)][f(i - 1, j - 1, k)] = 1;

    for(int i = 0; i <= 4; i++)
        for(int j = 0; j <= 4; j++)
            for(int k = 0; k < n; k++){
                if(i == 0 and j == 0 and k == 0)
                    for(int a = 0; a < n; a++)
                        for(int b = 0; b < n; b++){
                            int dl = min(sz(tab[a]), sz(tab[b]));
                            if(tab[a].substr(0, dl) == tab[b].substr(0, dl))
                                mat[f(0, 0, 0)][f(sz(tab[a]) - 1, sz(tab[b]) - 1,
                                    (sz(tab[a]) > sz(tab[b]) ? a : b))] += 1;
                        }
                else if(i != j and min(i, j) == 0)
                    for(int a = 0; a < n; a++){
                        int dl = min(abs(i - j), sz(tab[a]));

                        if(sz(tab[k]) - abs(i - j) < 0)
                            continue;
                        if(tab[a].substr(0, dl) ==
                           tab[k].substr(sz(tab[k]) - abs(i - j), dl))
                            mat[f(i, j, k)]
                               [f(i - 1 + (i < j) * sz(tab[a]), j - 1 + (j < i) * sz(tab[a]),
                                     (abs(i - j) > dl ? k : a))] += 1;
                    }
            }

    vvi pot = fast(mat, m);
    vvi ret = row * pot;

    cout << ret[0][f(0, 0, 0)] << '\n';
}

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

    int tt = 1;
    // cin >> tt;

    for(int i = 0; i < tt; i++)
        solve();
}