#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const long long nax = 2e5 + 10, mod = 998244353, inf = 1e18;
long long mul(long long x, long long y) { return x * y % mod; }
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<long long>> g(k), ad(k);
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    long long d;
    cin >> d;
    --d;
    for (long long j = 1; j < x; j++) {
      long long y;
      cin >> y;
      --y;
      g[d].push_back(y);
      ad[y].push_back(d);
      d = y;
    }
  }
  for (long long i = 0; i < k; i++) {
    sort(g[i].begin(), g[i].end());
    g[i].resize(unique(g[i].begin(), g[i].end()) - g[i].begin());
    sort(ad[i].begin(), ad[i].end());
    ad[i].resize(unique(ad[i].begin(), ad[i].end()) - ad[i].begin());
  }
  map<long long, long long> vis;
  vector<bool> used(k, 0), mark(k, 0);
  for (long long i = 0; i < k; i++) {
    if (mark[i]) continue;
    vector<long long> qq(1, i);
    mark[i] = 1;
    for (long long j = 0; j < qq.size(); j++) {
      for (auto it : g[qq[j]])
        if (!mark[it]) {
          qq.push_back(it);
          mark[it] = 1;
        }
      for (auto it : ad[qq[j]])
        if (!mark[it]) {
          qq.push_back(it);
          mark[it] = 1;
        }
    }
    long long val = 0;
    for (auto xx : qq) val += g[xx].size();
    if (val != qq.size() - 1) continue;
    long long e = -1;
    for (auto ii : qq)
      if (ad[ii].empty()) {
        e = ii;
        break;
      }
    vector<long long> seq(1, e);
    used[e] = 1;
    while (1) {
      long long nxt = -1;
      for (auto ii : g[seq.back()])
        if (!used[ii]) {
          nxt = ii;
          break;
        }
      if (nxt == -1) break;
      seq.push_back(nxt);
      used[nxt] = 1;
    }
    if (seq.size() == qq.size()) vis[seq.size()]++;
  }
  vector<pair<long long, long long>> et;
  for (auto it : vis) et.push_back(it);
  vector<long long> dp(m + 1);
  dp[0] = 1;
  for (long long i = 1; i < m + 1; i++) {
    for (auto it : et)
      if (i >= it.first) {
        dp[i] += mul(dp[i - it.first], it.second);
        dp[i] %= mod;
      }
  }
  cout << dp[m] << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long tt = 1;
  for (long long i = 1; i <= tt; i++) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const long long nax = 2e5 + 10, mod = 998244353, inf = 1e18;
long long mul(long long x, long long y) { return x * y % mod; }
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<long long>> g(k), ad(k);
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    long long d;
    cin >> d;
    --d;
    for (long long j = 1; j < x; j++) {
      long long y;
      cin >> y;
      --y;
      g[d].push_back(y);
      ad[y].push_back(d);
      d = y;
    }
  }
  for (long long i = 0; i < k; i++) {
    sort(g[i].begin(), g[i].end());
    g[i].resize(unique(g[i].begin(), g[i].end()) - g[i].begin());
    sort(ad[i].begin(), ad[i].end());
    ad[i].resize(unique(ad[i].begin(), ad[i].end()) - ad[i].begin());
  }
  map<long long, long long> vis;
  vector<bool> used(k, 0), mark(k, 0);
  for (long long i = 0; i < k; i++) {
    if (mark[i]) continue;
    vector<long long> qq(1, i);
    mark[i] = 1;
    for (long long j = 0; j < qq.size(); j++) {
      for (auto it : g[qq[j]])
        if (!mark[it]) {
          qq.push_back(it);
          mark[it] = 1;
        }
      for (auto it : ad[qq[j]])
        if (!mark[it]) {
          qq.push_back(it);
          mark[it] = 1;
        }
    }
    long long val = 0;
    for (auto xx : qq) val += g[xx].size();
    if (val != qq.size() - 1) continue;
    long long e = -1;
    for (auto ii : qq)
      if (ad[ii].empty()) {
        e = ii;
        break;
      }
    vector<long long> seq(1, e);
    used[e] = 1;
    while (1) {
      long long nxt = -1;
      for (auto ii : g[seq.back()])
        if (!used[ii]) {
          nxt = ii;
          break;
        }
      if (nxt == -1) break;
      seq.push_back(nxt);
      used[nxt] = 1;
    }
    if (seq.size() == qq.size()) vis[seq.size()]++;
  }
  vector<pair<long long, long long>> et;
  for (auto it : vis) et.push_back(it);
  vector<long long> dp(m + 1);
  dp[0] = 1;
  for (long long i = 1; i < m + 1; i++) {
    for (auto it : et)
      if (i >= it.first) {
        dp[i] += mul(dp[i - it.first], it.second);
        dp[i] %= mod;
      }
  }
  cout << dp[m] << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long tt = 1;
  for (long long i = 1; i <= tt; i++) {
    solve();
  }
  return 0;
}

```