#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10, mod = 998244353;
int n, m, K, fa[N], sz[N], pre[N], nex[N], a[N], b[N], tot;
int A[N], B[N], C[N], f[N];
bool vis[N], tag[N];
int find_(int x) { return fa[x] == x ? x : fa[x] = find_(fa[x]); }
void merge_(int u, int v) {
  u = find_(u), v = find_(v);
  if (u == v) return;
  if (sz[u] < sz[v]) swap(u, v);
  fa[v] = u, sz[u] += sz[v];
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= K; i++) fa[i] = i, sz[i] = 1;
  for (int i = 1, c; i <= n; i++) {
    scanf("%d", &c);
    for (int j = 1; j <= c; j++) {
      scanf("%d", &a[j]);
      merge_(a[j], a[1]);
      if (b[a[j]])
        vis[a[j]] = 1;
      else
        b[a[j]] = 1;
      if (j > 1) {
        if (pre[a[j]] && pre[a[j]] != a[j - 1])
          vis[a[j]] = 1;
        else
          pre[a[j]] = a[j - 1];
        if (nex[a[j - 1]] && nex[a[j - 1]] != a[j])
          vis[a[j - 1]] = 1;
        else
          nex[a[j - 1]] = a[j];
      }
    }
    for (int j = 1; j <= c; j++) b[a[j]] = 0;
  }
  for (int i = 1; i <= K; i++)
    vis[find_(i)] |= vis[i], tag[find_(i)] |= !nex[i];
  for (int i = 1; i <= K; i++)
    if (fa[i] == i && !vis[i] && tag[i]) {
      if (!A[sz[i]]) A[sz[i]] = ++tot, B[tot] = sz[i];
      C[A[sz[i]]]++;
    }
  f[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= tot; j++)
      if (B[j] <= i) f[i] = (f[i] + 1ll * f[i - B[j]] * C[j]) % mod;
  printf("%d\n", (f[m] % mod + mod) % mod);
}

### Prompt

Please create a solution in cpp 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 = 3e5 + 10, mod = 998244353;
int n, m, K, fa[N], sz[N], pre[N], nex[N], a[N], b[N], tot;
int A[N], B[N], C[N], f[N];
bool vis[N], tag[N];
int find_(int x) { return fa[x] == x ? x : fa[x] = find_(fa[x]); }
void merge_(int u, int v) {
  u = find_(u), v = find_(v);
  if (u == v) return;
  if (sz[u] < sz[v]) swap(u, v);
  fa[v] = u, sz[u] += sz[v];
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= K; i++) fa[i] = i, sz[i] = 1;
  for (int i = 1, c; i <= n; i++) {
    scanf("%d", &c);
    for (int j = 1; j <= c; j++) {
      scanf("%d", &a[j]);
      merge_(a[j], a[1]);
      if (b[a[j]])
        vis[a[j]] = 1;
      else
        b[a[j]] = 1;
      if (j > 1) {
        if (pre[a[j]] && pre[a[j]] != a[j - 1])
          vis[a[j]] = 1;
        else
          pre[a[j]] = a[j - 1];
        if (nex[a[j - 1]] && nex[a[j - 1]] != a[j])
          vis[a[j - 1]] = 1;
        else
          nex[a[j - 1]] = a[j];
      }
    }
    for (int j = 1; j <= c; j++) b[a[j]] = 0;
  }
  for (int i = 1; i <= K; i++)
    vis[find_(i)] |= vis[i], tag[find_(i)] |= !nex[i];
  for (int i = 1; i <= K; i++)
    if (fa[i] == i && !vis[i] && tag[i]) {
      if (!A[sz[i]]) A[sz[i]] = ++tot, B[tot] = sz[i];
      C[A[sz[i]]]++;
    }
  f[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= tot; j++)
      if (B[j] <= i) f[i] = (f[i] + 1ll * f[i - B[j]] * C[j]) % mod;
  printf("%d\n", (f[m] % mod + mod) % mod);
}

```