#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
using namespace std;
long long gcd(long long i, long long j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
template <typename T>
inline T getint() {
  T val = 0;
  char c;
  bool neg = false;
  while ((c = getchar()) && !(c >= '0' && c <= '9')) {
    neg |= c == '-';
  }
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val * (neg ? -1 : 1);
}
const long long INF = 3e18 + 100;
const int mod = 998244353;
const double eps = 1e-11, pi = acos(-1);
const long long maxN = 1 << 20, maxT = 600100, A = 179, K = 170;
mt19937 mt_rand(time(0));
long long bp(long long et, long long b) {
  b %= mod - 1;
  et %= mod;
  long long res = 1;
  for (int i = 30; i >= 0; --i) {
    res = (res * res) % mod;
    if ((b & (1 << i)) != 0) res = (res * et) % mod;
  }
  return res;
}
void panic() {
  cout << "-1\n";
  exit(0);
}
vector<vector<int>> g;
vector<int> used;
vector<set<int>> to(maxN);
int dfs(int v) {
  used[v] = 1;
  sort(g[v].begin(), g[v].end());
  g[v].resize(unique(g[v].begin(), g[v].end()) - g[v].begin());
  if (g[v].size() > 1) return -1;
  if (to[v].size() > 1) return -1;
  int l = 1;
  if (g[v].empty()) return l;
  if (used[g[v][0]]) return -1;
  int dd = dfs(g[v][0]);
  if (dd == -1) return -1;
  return l + dd;
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  g.assign(k, vector<int>());
  vector<int> d(k), bd(k);
  vector<vector<int>> tmp(n);
  for (int i = 0; i < n; ++i) {
    int c;
    cin >> c;
    tmp[i].resize(c);
    set<int> vert;
    for (auto& x : tmp[i]) {
      cin >> x;
      ++d[--x];
    }
    for (int j = 0; j + 1 < tmp[i].size(); ++j) {
      g[tmp[i][j]].push_back(tmp[i][j + 1]);
      to[tmp[i][j + 1]].insert(tmp[i][j]);
    }
  }
  used.assign(k, 0);
  vector<int> lns(k + 1);
  for (int i = 0; i < k; ++i) {
    if (d[i] == 0) {
      ++lns[1];
      continue;
    }
    if (!to[i].empty() || used[i]) continue;
    int tmp = dfs(i);
    if (tmp == -1) continue;
    ++lns[tmp];
  }
  vector<long long> dp(m + 1, 0);
  dp[0] = 1;
  set<int> s;
  for (int i = 0; i <= k; ++i) {
    if (lns[i] != 0) s.insert(i);
  }
  for (int i = 0; i < m; ++i) {
    for (auto x : s) {
      if (i + x > m) break;
      dp[i + x] = (dp[i + x] + dp[i] * lns[x]) % mod;
    }
  }
  cout << dp.back() << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  srand(time(0));
  int t = 1;
  while (t--) {
    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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
using namespace std;
long long gcd(long long i, long long j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
template <typename T>
inline T getint() {
  T val = 0;
  char c;
  bool neg = false;
  while ((c = getchar()) && !(c >= '0' && c <= '9')) {
    neg |= c == '-';
  }
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val * (neg ? -1 : 1);
}
const long long INF = 3e18 + 100;
const int mod = 998244353;
const double eps = 1e-11, pi = acos(-1);
const long long maxN = 1 << 20, maxT = 600100, A = 179, K = 170;
mt19937 mt_rand(time(0));
long long bp(long long et, long long b) {
  b %= mod - 1;
  et %= mod;
  long long res = 1;
  for (int i = 30; i >= 0; --i) {
    res = (res * res) % mod;
    if ((b & (1 << i)) != 0) res = (res * et) % mod;
  }
  return res;
}
void panic() {
  cout << "-1\n";
  exit(0);
}
vector<vector<int>> g;
vector<int> used;
vector<set<int>> to(maxN);
int dfs(int v) {
  used[v] = 1;
  sort(g[v].begin(), g[v].end());
  g[v].resize(unique(g[v].begin(), g[v].end()) - g[v].begin());
  if (g[v].size() > 1) return -1;
  if (to[v].size() > 1) return -1;
  int l = 1;
  if (g[v].empty()) return l;
  if (used[g[v][0]]) return -1;
  int dd = dfs(g[v][0]);
  if (dd == -1) return -1;
  return l + dd;
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  g.assign(k, vector<int>());
  vector<int> d(k), bd(k);
  vector<vector<int>> tmp(n);
  for (int i = 0; i < n; ++i) {
    int c;
    cin >> c;
    tmp[i].resize(c);
    set<int> vert;
    for (auto& x : tmp[i]) {
      cin >> x;
      ++d[--x];
    }
    for (int j = 0; j + 1 < tmp[i].size(); ++j) {
      g[tmp[i][j]].push_back(tmp[i][j + 1]);
      to[tmp[i][j + 1]].insert(tmp[i][j]);
    }
  }
  used.assign(k, 0);
  vector<int> lns(k + 1);
  for (int i = 0; i < k; ++i) {
    if (d[i] == 0) {
      ++lns[1];
      continue;
    }
    if (!to[i].empty() || used[i]) continue;
    int tmp = dfs(i);
    if (tmp == -1) continue;
    ++lns[tmp];
  }
  vector<long long> dp(m + 1, 0);
  dp[0] = 1;
  set<int> s;
  for (int i = 0; i <= k; ++i) {
    if (lns[i] != 0) s.insert(i);
  }
  for (int i = 0; i < m; ++i) {
    for (auto x : s) {
      if (i + x > m) break;
      dp[i + x] = (dp[i + x] + dp[i] * lns[x]) % mod;
    }
  }
  cout << dp.back() << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  srand(time(0));
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```