#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
const int LOG2 = 18;
const int MOD = 998244353;
const long long oo = 1e18;
set<int> g[N];
vector<int> g2[N];
int in[N], seen[N], sz, bad[N], val[N];
void mark_bad(int u) {
  bad[u] = 1;
  for (int v : g2[u])
    if (!bad[v]) mark_bad(v);
}
int go(int u) {
  int ok = 1;
  sz++;
  seen[u] = 1;
  ok &= g[u].size() <= 1;
  for (int v : g[u]) {
    if (!seen[v])
      ok &= go(v);
    else
      ok = 0;
  }
  return ok;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  for (int it = 0; it < n; it++) {
    int len;
    cin >> len;
    int last = -1;
    while (len--) {
      int x;
      cin >> x;
      if (last != -1) {
        in[x]++;
        g[last].insert(x);
        g2[last].push_back(x);
        g2[x].push_back(last);
      }
      last = x;
    }
  }
  map<int, int> by_sz;
  for (int i = 1; i <= k; i++) {
    if (in[i] == 0 && !bad[i]) {
      sz = 0;
      if (!go(i)) mark_bad(i);
      val[i] = sz;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!seen[i]) {
      if (!go(i)) mark_bad(i);
    }
  }
  for (int i = 1; i <= k; i++) {
    if (in[i] == 0 && !bad[i]) {
      by_sz[val[i]]++;
    }
  }
  vector<int> dp(2 * N + 2);
  dp[m] = 1;
  for (int i = m - 1; i >= 0; i--) {
    for (auto e : by_sz) {
      dp[i] += 1ll * dp[i + e.first] * e.second % MOD;
      if (dp[i] >= MOD) dp[i] -= MOD;
    }
  }
  cout << dp[0] << '\n';
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 int N = 3e5 + 10;
const int LOG2 = 18;
const int MOD = 998244353;
const long long oo = 1e18;
set<int> g[N];
vector<int> g2[N];
int in[N], seen[N], sz, bad[N], val[N];
void mark_bad(int u) {
  bad[u] = 1;
  for (int v : g2[u])
    if (!bad[v]) mark_bad(v);
}
int go(int u) {
  int ok = 1;
  sz++;
  seen[u] = 1;
  ok &= g[u].size() <= 1;
  for (int v : g[u]) {
    if (!seen[v])
      ok &= go(v);
    else
      ok = 0;
  }
  return ok;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  for (int it = 0; it < n; it++) {
    int len;
    cin >> len;
    int last = -1;
    while (len--) {
      int x;
      cin >> x;
      if (last != -1) {
        in[x]++;
        g[last].insert(x);
        g2[last].push_back(x);
        g2[x].push_back(last);
      }
      last = x;
    }
  }
  map<int, int> by_sz;
  for (int i = 1; i <= k; i++) {
    if (in[i] == 0 && !bad[i]) {
      sz = 0;
      if (!go(i)) mark_bad(i);
      val[i] = sz;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!seen[i]) {
      if (!go(i)) mark_bad(i);
    }
  }
  for (int i = 1; i <= k; i++) {
    if (in[i] == 0 && !bad[i]) {
      by_sz[val[i]]++;
    }
  }
  vector<int> dp(2 * N + 2);
  dp[m] = 1;
  for (int i = m - 1; i >= 0; i--) {
    for (auto e : by_sz) {
      dp[i] += 1ll * dp[i + e.first] * e.second % MOD;
      if (dp[i] >= MOD) dp[i] -= MOD;
    }
  }
  cout << dp[0] << '\n';
}

```