#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10, mod = 998244353;
int n, m, k, a[N], fa[N], nxt[N], pre[N], tg[N], sz[N], ct[N], stk[N], f[N];
inline int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
inline void merge(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) {
    tg[x] = 1;
    return;
  }
  fa[x] = y;
  sz[y] += sz[x];
  tg[y] |= tg[x];
}
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]);
      if (j == 1) continue;
      if (nxt[a[j - 1]] && nxt[a[j - 1]] != a[j])
        tg[find(a[j - 1])] = tg[find(a[j])] = 1;
      else if (pre[a[j]] && pre[a[j]] != a[j - 1])
        tg[find(a[j - 1])] = tg[find(a[j])] = 1;
      else if (!nxt[a[j - 1]])
        nxt[a[j - 1]] = a[j], pre[a[j]] = a[j - 1], merge(a[j - 1], a[j]);
    }
  }
  for (int i = 1; i <= k; ++i)
    if (find(i) == i && !tg[i]) ct[sz[i]]++;
  int top = 0;
  for (int i = 1; i <= k; ++i)
    if (ct[i]) stk[++top] = i;
  f[0] = 1;
  for (int i = 0; i <= m; ++i)
    for (int j = 1; j <= top && i + stk[j] <= m; ++j)
      f[i + stk[j]] = (f[i + stk[j]] + 1ll * f[i] * ct[stk[j]]) % mod;
  printf("%d\n", f[m]);
  return 0;
}

### 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>
using namespace std;
const int N = 3e5 + 10, mod = 998244353;
int n, m, k, a[N], fa[N], nxt[N], pre[N], tg[N], sz[N], ct[N], stk[N], f[N];
inline int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
inline void merge(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) {
    tg[x] = 1;
    return;
  }
  fa[x] = y;
  sz[y] += sz[x];
  tg[y] |= tg[x];
}
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]);
      if (j == 1) continue;
      if (nxt[a[j - 1]] && nxt[a[j - 1]] != a[j])
        tg[find(a[j - 1])] = tg[find(a[j])] = 1;
      else if (pre[a[j]] && pre[a[j]] != a[j - 1])
        tg[find(a[j - 1])] = tg[find(a[j])] = 1;
      else if (!nxt[a[j - 1]])
        nxt[a[j - 1]] = a[j], pre[a[j]] = a[j - 1], merge(a[j - 1], a[j]);
    }
  }
  for (int i = 1; i <= k; ++i)
    if (find(i) == i && !tg[i]) ct[sz[i]]++;
  int top = 0;
  for (int i = 1; i <= k; ++i)
    if (ct[i]) stk[++top] = i;
  f[0] = 1;
  for (int i = 0; i <= m; ++i)
    for (int j = 1; j <= top && i + stk[j] <= m; ++j)
      f[i + stk[j]] = (f[i + stk[j]] + 1ll * f[i] * ct[stk[j]]) % mod;
  printf("%d\n", f[m]);
  return 0;
}

```