A subarray of array a from index l to the index r is the array [a_l, a_{l+1}, ..., a_{r}]. The number of occurrences of the array b in the array a is the number of subarrays of a such that they are equal to b.

You are given n arrays A_1, A_2, ..., A_n; the elements of these arrays are integers from 1 to k. You have to build an array a consisting of m integers from 1 to k in such a way that, for every given subarray A_i, the number of occurrences of A_i in the array a is not less than the number of occurrences of each non-empty subarray of A_i in a. Note that if A_i doesn't occur in a, and no subarray of A_i occurs in a, this condition is still met for A_i.

Your task is to calculate the number of different arrays a you can build, and print it modulo 998244353.

Input

The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 3 ⋅ 10^5) — the number of the given arrays, the desired length of the array a, and the upper bound on the values in the arrays.

Then n lines follow. The i-th line represents the array A_i. The first integer in the i-th line is c_i (1 ≤ c_i ≤ m) — the number of elements in A_i; then, c_i integers from 1 to k follow — the elements of the array A_i.

Additional constraint on the input: ∑_{i=1}^n c_i ≤ 3 ⋅ 10^5; i. e., the number of elements in the given arrays in total does not exceed 3 ⋅ 10^5.

Output

Print one integer — the number of different arrays a you can build, taken modulo 998244353.

Examples

Input


2 4 3
2 1 2
1 3


Output


5


Input


2 4 3
2 1 2
3 3 2 1


Output


0


Input


1 42 1337
2 13 31


Output


721234447
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <iomanip>
#include <cstdlib>
#include <fstream>
#include <ctime>
#include <random>

#define int long long

using namespace std;

const int INF = 998244353;
int in[300005], dp[300005];
bool used[300005];
vector < int > ch, edge[300005];
vector < vector < int > > a;
vector < pair < int, int > > cnt;

pair < bool, int > dfs(int v)
{
    if(in[v] > 1 || (int)edge[v].size() > 1 || used[v] == 1)
        return {0, 0};
    used[v] = 1;
    if(edge[v].empty())
        return {1, 1};
    pair < bool, int > p = dfs(edge[v][0]);
    p.second++;
    return p;
}

main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, m, k;
    cin >> n >> m >> k;
    a.resize(n);
    for(int i = 0; i < n; i++)
    {
        int sz;
        cin >> sz;
        a[i].resize(sz);
        for(int j = 0; j < sz; j++)
            cin >> a[i][j];
    }
    sort(a.begin(), a.end());
    a.erase(unique(a.begin(), a.end()), a.end());
    n = a.size();
    for(int i = 0; i < n; i++)
        for(int j = 0; j < (int)a[i].size() - 1; j++)
        {
            edge[a[i][j]].push_back(a[i][j + 1]);
            in[a[i][j + 1]]++;
        }
    for(int i = 1; i <= k; i++)
        if(!in[i])
        {
            pair < bool, int > p = dfs(i);
            if(p.first)
                ch.push_back(p.second);
        }
    sort(ch.begin(), ch.end());
    if(ch.size())
        cnt.push_back({ch[0], 1});
    for(int i = 1; i < (int)ch.size(); i++)
        if(ch[i] == ch[i - 1])
            cnt.back().second++;
        else
            cnt.push_back({ch[i], 1});
    dp[0] = 1;
    for(int i = 1; i <= m; i++)
    {
        for(auto j : cnt)
            if(j.first <= i)
                dp[i] = (dp[i] + dp[i - j.first] * j.second) % INF;
    }
    cout << dp[m];
    return 0;
}
