#include <bits/stdc++.h>
using namespace std;
const int N = 300500;
const int L = 22;
const long long M = 998244353;
int par[N];
int up[N][L];
int sz[N];
void merge(int a, int b) {
  if (sz[a] <= sz[b]) {
    par[a] = b;
    sz[b] += sz[a];
  } else {
    par[b] = a;
    sz[a] += sz[b];
  }
}
int parent(int a) {
  if (a == par[a]) return a;
  return par[a] = parent(par[a]);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  for (int i = 0; i < N; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<int>> A;
  for (int _ = 0; _ < n; _++) {
    int l;
    cin >> l;
    vector<int> B(l);
    for (int i = 0; i < l; i++) {
      cin >> B[i];
    }
    A.push_back(B);
  }
  for (auto B : A) {
    for (int i = 0; i < (int)B.size() - 1; i++) {
      merge(parent(B[i]), parent(B[i + 1]));
    }
  }
  vector<bool> ban(k + 1, false);
  vector<int> next(k + 1, -1);
  vector<int> prev(k + 1, -1);
  for (auto B : A) {
    for (int i = 0; i < (int)B.size() - 1; i++) {
      if (next[B[i]] == -1) {
        next[B[i]] = B[i + 1];
      } else if (next[B[i]] != B[i + 1]) {
        ban[parent(B[i])] = true;
      }
    }
    for (int i = 0; i < (int)B.size() - 1; i++) {
      if (prev[B[i + 1]] == -1) {
        prev[B[i + 1]] = B[i];
      } else if (prev[B[i + 1]] != B[i]) {
        ban[parent(B[i + 1])] = true;
      }
    }
    vector<int> B2 = B;
    sort(B2.begin(), B2.end());
    for (int i = 0; i < (int)B2.size() - 1; i++) {
      if (B2[i] == B2[i + 1]) {
        ban[parent(B2[i])] = true;
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    if (ban[parent(i)]) {
      ban[i] = true;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!ban[i]) {
      if (next[i] != -1) {
        up[i][0] = next[i];
      } else {
        up[i][0] = i;
      }
    }
  }
  for (int l = 1; l < L; l++) {
    for (int i = 1; i <= k; i++) {
      if (!ban[i]) {
        up[i][l] = up[up[i][l - 1]][l - 1];
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!ban[i]) {
      if (next[up[i][L - 1]] != -1) {
        ban[i] = true;
      }
    }
  }
  vector<int> aboba;
  for (int i = 1; i <= k; i++) {
    if (!ban[i] && prev[i] == -1) {
      int j = i;
      int kek = 1;
      for (int l = L - 1; l >= 0; l--) {
        if (next[up[j][l]] != -1) {
          j = up[j][l];
          kek += (1 << l);
        }
      }
      if (next[j] != -1) {
        kek++;
      }
      aboba.push_back(kek);
    }
  }
  vector<int> boba;
  int s = (int)sqrt(n);
  vector<long long> cnt(s + 1);
  for (auto x : aboba) {
    if (x <= s) {
      cnt[x]++;
    } else {
      boba.push_back(x);
    }
  }
  vector<long long> dp(m + 1);
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int x = 1; x <= min(i, s); x++) {
      dp[i] += (cnt[x] * dp[i - x]) % M;
    }
    for (auto x : boba) {
      if (i >= x) {
        dp[i] += dp[i - x];
      }
    }
    dp[i] %= M;
  }
  cout << dp[m] << '\n';
}

### 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 int N = 300500;
const int L = 22;
const long long M = 998244353;
int par[N];
int up[N][L];
int sz[N];
void merge(int a, int b) {
  if (sz[a] <= sz[b]) {
    par[a] = b;
    sz[b] += sz[a];
  } else {
    par[b] = a;
    sz[a] += sz[b];
  }
}
int parent(int a) {
  if (a == par[a]) return a;
  return par[a] = parent(par[a]);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  for (int i = 0; i < N; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<int>> A;
  for (int _ = 0; _ < n; _++) {
    int l;
    cin >> l;
    vector<int> B(l);
    for (int i = 0; i < l; i++) {
      cin >> B[i];
    }
    A.push_back(B);
  }
  for (auto B : A) {
    for (int i = 0; i < (int)B.size() - 1; i++) {
      merge(parent(B[i]), parent(B[i + 1]));
    }
  }
  vector<bool> ban(k + 1, false);
  vector<int> next(k + 1, -1);
  vector<int> prev(k + 1, -1);
  for (auto B : A) {
    for (int i = 0; i < (int)B.size() - 1; i++) {
      if (next[B[i]] == -1) {
        next[B[i]] = B[i + 1];
      } else if (next[B[i]] != B[i + 1]) {
        ban[parent(B[i])] = true;
      }
    }
    for (int i = 0; i < (int)B.size() - 1; i++) {
      if (prev[B[i + 1]] == -1) {
        prev[B[i + 1]] = B[i];
      } else if (prev[B[i + 1]] != B[i]) {
        ban[parent(B[i + 1])] = true;
      }
    }
    vector<int> B2 = B;
    sort(B2.begin(), B2.end());
    for (int i = 0; i < (int)B2.size() - 1; i++) {
      if (B2[i] == B2[i + 1]) {
        ban[parent(B2[i])] = true;
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    if (ban[parent(i)]) {
      ban[i] = true;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!ban[i]) {
      if (next[i] != -1) {
        up[i][0] = next[i];
      } else {
        up[i][0] = i;
      }
    }
  }
  for (int l = 1; l < L; l++) {
    for (int i = 1; i <= k; i++) {
      if (!ban[i]) {
        up[i][l] = up[up[i][l - 1]][l - 1];
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!ban[i]) {
      if (next[up[i][L - 1]] != -1) {
        ban[i] = true;
      }
    }
  }
  vector<int> aboba;
  for (int i = 1; i <= k; i++) {
    if (!ban[i] && prev[i] == -1) {
      int j = i;
      int kek = 1;
      for (int l = L - 1; l >= 0; l--) {
        if (next[up[j][l]] != -1) {
          j = up[j][l];
          kek += (1 << l);
        }
      }
      if (next[j] != -1) {
        kek++;
      }
      aboba.push_back(kek);
    }
  }
  vector<int> boba;
  int s = (int)sqrt(n);
  vector<long long> cnt(s + 1);
  for (auto x : aboba) {
    if (x <= s) {
      cnt[x]++;
    } else {
      boba.push_back(x);
    }
  }
  vector<long long> dp(m + 1);
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int x = 1; x <= min(i, s); x++) {
      dp[i] += (cnt[x] * dp[i - x]) % M;
    }
    for (auto x : boba) {
      if (i >= x) {
        dp[i] += dp[i - x];
      }
    }
    dp[i] %= M;
  }
  cout << dp[m] << '\n';
}

```