#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300005;
const int Mod = 998244353;
vector<int> pre[MAXN], nxt[MAXN];
bool vis[MAXN];
void dfs(int u, int las, int& ok, int& siz) {
  vis[u] = 1;
  ++siz;
  ok &= (pre[u].size() <= 1u);
  ok &= (nxt[u].size() <= 1u);
  for (auto& v : pre[u]) {
    if (v == las) continue;
    if (vis[v]) {
      ok = 0;
      continue;
    }
    dfs(v, u, ok, siz);
  }
  for (auto& v : nxt[u]) {
    if (v == las) continue;
    if (vis[v]) {
      ok = 0;
      continue;
    }
    dfs(v, u, ok, siz);
  }
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    int c;
    scanf("%d", &c);
    vector<int> a(c);
    for (int j = 0; j < c; j++) scanf("%d", &a[j]);
    for (int j = 0; j + 1 < c; j++) {
      nxt[a[j]].push_back(a[j + 1]);
      pre[a[j + 1]].push_back(a[j]);
    }
  }
  for (int i = 1; i <= k; i++) {
    sort(nxt[i].begin(), nxt[i].end());
    nxt[i].erase(unique(nxt[i].begin(), nxt[i].end()), nxt[i].end());
    sort(pre[i].begin(), pre[i].end());
    pre[i].erase(unique(pre[i].begin(), pre[i].end()), pre[i].end());
  }
  map<int, int> mp;
  for (int i = 1; i <= k; i++) {
    if (vis[i]) continue;
    int ok = 1, siz = 0;
    dfs(i, 0, ok, siz);
    if (ok) mp[siz]++;
  }
  static int dp[MAXN];
  for (int i = (dp[0] = 1); i <= m; i++)
    for (auto& t : mp) {
      int j = t.first, c = t.second;
      if (j <= i) dp[i] = (dp[i] + 1LL * c * dp[i - j]) % Mod;
    }
  printf("%d\n", dp[m]);
  return 0;
}

### Prompt

Develop a solution in cpp 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 MAXN = 300005;
const int Mod = 998244353;
vector<int> pre[MAXN], nxt[MAXN];
bool vis[MAXN];
void dfs(int u, int las, int& ok, int& siz) {
  vis[u] = 1;
  ++siz;
  ok &= (pre[u].size() <= 1u);
  ok &= (nxt[u].size() <= 1u);
  for (auto& v : pre[u]) {
    if (v == las) continue;
    if (vis[v]) {
      ok = 0;
      continue;
    }
    dfs(v, u, ok, siz);
  }
  for (auto& v : nxt[u]) {
    if (v == las) continue;
    if (vis[v]) {
      ok = 0;
      continue;
    }
    dfs(v, u, ok, siz);
  }
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    int c;
    scanf("%d", &c);
    vector<int> a(c);
    for (int j = 0; j < c; j++) scanf("%d", &a[j]);
    for (int j = 0; j + 1 < c; j++) {
      nxt[a[j]].push_back(a[j + 1]);
      pre[a[j + 1]].push_back(a[j]);
    }
  }
  for (int i = 1; i <= k; i++) {
    sort(nxt[i].begin(), nxt[i].end());
    nxt[i].erase(unique(nxt[i].begin(), nxt[i].end()), nxt[i].end());
    sort(pre[i].begin(), pre[i].end());
    pre[i].erase(unique(pre[i].begin(), pre[i].end()), pre[i].end());
  }
  map<int, int> mp;
  for (int i = 1; i <= k; i++) {
    if (vis[i]) continue;
    int ok = 1, siz = 0;
    dfs(i, 0, ok, siz);
    if (ok) mp[siz]++;
  }
  static int dp[MAXN];
  for (int i = (dp[0] = 1); i <= m; i++)
    for (auto& t : mp) {
      int j = t.first, c = t.second;
      if (j <= i) dp[i] = (dp[i] + 1LL * c * dp[i - j]) % Mod;
    }
  printf("%d\n", dp[m]);
  return 0;
}

```