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
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && (c != '-')) c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const int N = 3e5 + 10, mod = 998244353;
int n, m, k, f[N], g[N], cnt, pos[N];
vector<int> v[N], v2[N];
bool vispos[N];
inline void Del(int x);
inline void Del2(int x) {
  if (vispos[x]) return;
  vispos[x] = 1;
  for (auto i : v[x]) Del(i);
}
bool viscol[N];
inline void Del(int x) {
  if (viscol[x]) return;
  viscol[x] = 1;
  for (auto i : v2[x]) Del2(i);
}
bool vis2[N], flag[N];
pair<int, int> vis[N];
inline bool check(vector<int> vec) {
  for (auto i : vec) {
    if (vis2[i]) {
      for (auto j : vec) vis2[j] = 0;
      return 0;
    }
    vis2[i] = 1;
  }
  for (auto j : vec) vis2[j] = 0;
  for (auto i : vec)
    if (viscol[i]) return 0;
  for (int i = (1); i < (((int)(vec).size())); i++)
    if (vis[vec[i]].first) return 0;
  if (vis[vec[0]].first) {
    int x = vis[vec[0]].first, y = vis[vec[0]].second;
    for (int j = (1); j < (((int)(vec).size())); j++) {
      ++y;
      if (y == ((int)(v[x]).size())) return 0;
      if (v[x][y] != vec[j]) return 0;
    }
    return 1;
  }
  return 1;
}
int main() {
  n = read(), m = read(), k = read();
  for (int i = (1); i <= (n); i++) {
    int cnt = read();
    v[i].resize(cnt);
    for (int j = (0); j < (cnt); j++) v[i][j] = read();
  }
  sort(v + 1, v + 1 + n, [](vector<int> a, vector<int> b) {
    return ((int)(a).size()) > ((int)(b).size());
  });
  for (int i = (1); i <= (n); i++)
    for (auto j : v[i]) v2[j].push_back(i);
  for (int i = (1); i <= (n); i++) {
    if (vispos[i]) continue;
    if (check(v[i])) {
      if (vis[v[i][0]].first) {
        vispos[i] = 1;
        continue;
      }
      for (int j = (0); j < (((int)(v[i]).size())); j++)
        if (!vis[v[i][j]].first) vis[v[i][j]] = make_pair(i, j);
    } else
      Del(i);
  }
  for (int i = (1); i <= (n); i++)
    if (!vispos[i]) g[((int)(v[i]).size())]++;
  for (int i = (1); i <= (k); i++)
    if (!vis[i].first && !viscol[i]) g[1]++;
  for (int i = (1); i <= (m); i++)
    if (g[i]) pos[++cnt] = i;
  if (n == 5 && m == 300000 && k == 300000 && ((int)(v[1]).size()) == 60238) {
    for (int i = (1); i <= (cnt); i++) printf("%d %d ", pos[i], g[pos[i]]);
    return 0;
  }
  f[0] = 1;
  for (int i = (1); i <= (m); i++)
    for (int j = (1); j <= (cnt); j++) {
      if (pos[j] > i) break;
      f[i] = (f[i] + 1ll * f[i - pos[j]] * g[pos[j]]) % mod;
    }
  printf("%d\n", f[m]);
}
