#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int N = 3e5 + 9, mod = 998244353;
int n, m, k;
set<int> nxt[N], prv[N];
bool bad[N];
void dfsbads(int node) {
  if (bad[node]) return;
  bad[node] = true;
  for (auto i : nxt[node]) dfsbads(i);
  for (auto i : prv[node]) dfsbads(i);
}
int dfsChain(int node) {
  if (nxt[node].empty()) return 1;
  assert(nxt[node].size() == 1);
  return 1 + dfsChain(*nxt[node].begin());
}
int freq[N];
int dp[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  while (n--) {
    int c, x;
    cin >> c >> x;
    for (int i = 1; i < c; i++) {
      int y;
      cin >> y;
      nxt[x].insert(y);
      prv[y].insert(x);
      x = y;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (nxt[i].size() > 1 || prv[i].size() > 1) dfsbads(i);
  }
  for (int i = 1; i <= k; i++) {
    if (bad[i] || prv[i].size() > 0) continue;
    freq[dfsChain(i)]++;
  }
  vector<pair<int, int> > v;
  for (int i = 1; i <= m; i++) {
    if (freq[i]) {
      v.push_back({i, freq[i]});
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (auto j : v) {
      if (j.first > i) break;
      dp[i] = (dp[i] + 1LL * j.second * dp[i - j.first]) % mod;
    }
  }
  cout << dp[m];
}

### Prompt

Your task is to create 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("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int N = 3e5 + 9, mod = 998244353;
int n, m, k;
set<int> nxt[N], prv[N];
bool bad[N];
void dfsbads(int node) {
  if (bad[node]) return;
  bad[node] = true;
  for (auto i : nxt[node]) dfsbads(i);
  for (auto i : prv[node]) dfsbads(i);
}
int dfsChain(int node) {
  if (nxt[node].empty()) return 1;
  assert(nxt[node].size() == 1);
  return 1 + dfsChain(*nxt[node].begin());
}
int freq[N];
int dp[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  while (n--) {
    int c, x;
    cin >> c >> x;
    for (int i = 1; i < c; i++) {
      int y;
      cin >> y;
      nxt[x].insert(y);
      prv[y].insert(x);
      x = y;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (nxt[i].size() > 1 || prv[i].size() > 1) dfsbads(i);
  }
  for (int i = 1; i <= k; i++) {
    if (bad[i] || prv[i].size() > 0) continue;
    freq[dfsChain(i)]++;
  }
  vector<pair<int, int> > v;
  for (int i = 1; i <= m; i++) {
    if (freq[i]) {
      v.push_back({i, freq[i]});
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (auto j : v) {
      if (j.first > i) break;
      dp[i] = (dp[i] + 1LL * j.second * dp[i - j.first]) % mod;
    }
  }
  cout << dp[m];
}

```