#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353, N = 3e5 + 5;
pair<int, int> g[N];
int nxt[N], pre[N], a[N], ban[N], vis[N], t[N], dp[N];
int n, m, k, len, flag;
void link(int u, int v) {
  if (nxt[u] && nxt[u] != v) {
    flag = 1;
    return;
  }
  if (pre[v] && pre[v] != u) {
    flag = 1;
    return;
  }
  nxt[u] = v;
  pre[v] = u;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    int len;
    scanf("%d", &len);
    flag = 0;
    for (int j = 1; j <= len; j++) {
      scanf("%d", &a[j]);
      if (j > 1) link(a[j - 1], a[j]);
    }
    if (flag)
      for (int j = 1; j <= len; j++) ban[a[j]] = 1;
  }
  for (int i = 1; i <= k; i++) {
    if (ban[i]) continue;
    if (!pre[i]) {
      int tmp = i, cando = 1;
      while (tmp) {
        if (ban[tmp]) cando = 0;
        tmp = nxt[tmp];
      }
      if (cando) {
        int qwq = 0;
        tmp = i;
        while (tmp) qwq++, vis[tmp] = 1, tmp = nxt[tmp];
        t[qwq]++;
      }
    }
  }
  for (int i = 1; i <= k; i++)
    if (t[i]) g[++len] = make_pair(i, t[i]);
  dp[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= len; j++) {
      if (g[j].first > i) break;
      dp[i] = (dp[i] + 1ll * dp[i - g[j].first] * g[j].second % mod) % mod;
    }
  printf("%d\n", dp[m]);
  return 0;
}

### Prompt

Your challenge is to write 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 mod = 998244353, N = 3e5 + 5;
pair<int, int> g[N];
int nxt[N], pre[N], a[N], ban[N], vis[N], t[N], dp[N];
int n, m, k, len, flag;
void link(int u, int v) {
  if (nxt[u] && nxt[u] != v) {
    flag = 1;
    return;
  }
  if (pre[v] && pre[v] != u) {
    flag = 1;
    return;
  }
  nxt[u] = v;
  pre[v] = u;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    int len;
    scanf("%d", &len);
    flag = 0;
    for (int j = 1; j <= len; j++) {
      scanf("%d", &a[j]);
      if (j > 1) link(a[j - 1], a[j]);
    }
    if (flag)
      for (int j = 1; j <= len; j++) ban[a[j]] = 1;
  }
  for (int i = 1; i <= k; i++) {
    if (ban[i]) continue;
    if (!pre[i]) {
      int tmp = i, cando = 1;
      while (tmp) {
        if (ban[tmp]) cando = 0;
        tmp = nxt[tmp];
      }
      if (cando) {
        int qwq = 0;
        tmp = i;
        while (tmp) qwq++, vis[tmp] = 1, tmp = nxt[tmp];
        t[qwq]++;
      }
    }
  }
  for (int i = 1; i <= k; i++)
    if (t[i]) g[++len] = make_pair(i, t[i]);
  dp[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= len; j++) {
      if (g[j].first > i) break;
      dp[i] = (dp[i] + 1ll * dp[i - g[j].first] * g[j].second % mod) % mod;
    }
  printf("%d\n", dp[m]);
  return 0;
}

```