#include <bits/stdc++.h>
using namespace std;
const int c = 300005;
int n, m, k, db[c], el[c], kov[c], ki[c];
vector<int> sz[c];
bool rossz[c], v[c];
vector<pair<int, long long> > lep;
long long dp[c], mod = 998244353;
void dfs(int a) {
  v[a] = true;
  rossz[a] = 1;
  for (auto x : sz[a]) {
    if (!v[x]) {
      dfs(x);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    int h, pr;
    cin >> h;
    cin >> pr;
    ki[pr] = i;
    for (int j = 1; j < h; j++) {
      int x;
      cin >> x;
      sz[x].push_back(pr);
      sz[pr].push_back(x);
      if (ki[x] == i) {
        rossz[x] = 1;
      }
      ki[x] = i;
      if (kov[pr] && kov[pr] != x) {
        rossz[pr] = 1;
      }
      if (el[x] && el[x] != pr) {
        rossz[x] = 1;
      }
      kov[pr] = x;
      el[x] = pr;
      pr = x;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (rossz[i] && !v[i]) {
      dfs(i);
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!rossz[i] && !el[i]) {
      int len = 1, pos = i;
      while (kov[pos]) {
        pos = kov[pos];
        len++;
      }
      db[len]++;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (db[i]) {
      lep.push_back({i, db[i]});
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (auto x : lep) {
      if (i >= x.first) {
        dp[i] += dp[i - x.first] * x.second;
      }
    }
    dp[i] %= mod;
  }
  cout << dp[m] << "\n";
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 c = 300005;
int n, m, k, db[c], el[c], kov[c], ki[c];
vector<int> sz[c];
bool rossz[c], v[c];
vector<pair<int, long long> > lep;
long long dp[c], mod = 998244353;
void dfs(int a) {
  v[a] = true;
  rossz[a] = 1;
  for (auto x : sz[a]) {
    if (!v[x]) {
      dfs(x);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    int h, pr;
    cin >> h;
    cin >> pr;
    ki[pr] = i;
    for (int j = 1; j < h; j++) {
      int x;
      cin >> x;
      sz[x].push_back(pr);
      sz[pr].push_back(x);
      if (ki[x] == i) {
        rossz[x] = 1;
      }
      ki[x] = i;
      if (kov[pr] && kov[pr] != x) {
        rossz[pr] = 1;
      }
      if (el[x] && el[x] != pr) {
        rossz[x] = 1;
      }
      kov[pr] = x;
      el[x] = pr;
      pr = x;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (rossz[i] && !v[i]) {
      dfs(i);
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!rossz[i] && !el[i]) {
      int len = 1, pos = i;
      while (kov[pos]) {
        pos = kov[pos];
        len++;
      }
      db[len]++;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (db[i]) {
      lep.push_back({i, db[i]});
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (auto x : lep) {
      if (i >= x.first) {
        dp[i] += dp[i - x.first] * x.second;
      }
    }
    dp[i] %= mod;
  }
  cout << dp[m] << "\n";
  return 0;
}

```