#include <bits/stdc++.h>
using namespace std;
const int maxn = 300005, mod = 998244353;
int n, m, k, mx;
int in[maxn], out[maxn], f[maxn], tot[maxn], q[maxn], tmp[maxn];
vector<int> t, v[maxn], g[maxn];
map<int, int> mp[maxn];
void dfs(int x, int d) {
  if (in[x] > 1 || out[x] > 1) return;
  if (out[x] == 0) tot[d]++, mx = max(mx, d);
  for (int i = 0; i < g[x].size(); i++) dfs(g[x][i], d + 1);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, c; i <= n; i++) {
    scanf("%d", &c), v[i].resize(c);
    for (int j = 0; j < c; j++) {
      scanf("%d", &v[i][j]);
      if (j > 0 && mp[v[i][j - 1]][v[i][j]] == 0)
        mp[v[i][j - 1]][v[i][j]] = 1, g[v[i][j - 1]].push_back(v[i][j]),
                    out[v[i][j - 1]]++, in[v[i][j]]++;
    }
  }
  for (int i = 1; i <= k; i++)
    if (in[i] == 0) dfs(i, 1);
  f[0] = 1;
  for (int i = 1; i <= mx; i++)
    if (tot[i]) t.push_back(i);
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < t.size(); j++)
      if (i >= t[j]) f[i] = (f[i] + 1ll * f[i - t[j]] * tot[t[j]]) % mod;
  printf("%d\n", f[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, mod = 998244353;
int n, m, k, mx;
int in[maxn], out[maxn], f[maxn], tot[maxn], q[maxn], tmp[maxn];
vector<int> t, v[maxn], g[maxn];
map<int, int> mp[maxn];
void dfs(int x, int d) {
  if (in[x] > 1 || out[x] > 1) return;
  if (out[x] == 0) tot[d]++, mx = max(mx, d);
  for (int i = 0; i < g[x].size(); i++) dfs(g[x][i], d + 1);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, c; i <= n; i++) {
    scanf("%d", &c), v[i].resize(c);
    for (int j = 0; j < c; j++) {
      scanf("%d", &v[i][j]);
      if (j > 0 && mp[v[i][j - 1]][v[i][j]] == 0)
        mp[v[i][j - 1]][v[i][j]] = 1, g[v[i][j - 1]].push_back(v[i][j]),
                    out[v[i][j - 1]]++, in[v[i][j]]++;
    }
  }
  for (int i = 1; i <= k; i++)
    if (in[i] == 0) dfs(i, 1);
  f[0] = 1;
  for (int i = 1; i <= mx; i++)
    if (tot[i]) t.push_back(i);
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < t.size(); j++)
      if (i >= t[j]) f[i] = (f[i] + 1ll * f[i - t[j]] * tot[t[j]]) % mod;
  printf("%d\n", f[m]);
  return 0;
}

```