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;
int n, m, k;
vector<int> g[300010], con;
vector<int> unz;
int d[300010], vis[300010];
int a[300010];
int f[300010];
void dfs(int u) {
  vis[u] = true, con.push_back(u);
  for (int v : g[u])
    if (!vis[v]) dfs(v);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    int c;
    scanf("%d", &c);
    int pre = -1, now = -1;
    while (c--) {
      scanf("%d", &now);
      if (pre != -1)
        g[pre].push_back(now), g[now].push_back(pre), d[pre]++, d[now]++;
      pre = now;
    }
  }
  for (int i = 1; i <= k; i++)
    if (!vis[i]) {
      con.clear();
      dfs(i);
      int c1 = 0, c2 = 0;
      for (int x : con) c1 += (d[x] == 1), c2 += (d[x] == 2);
      if ((con.size() == 1 && c1 == 1) || (c1 + c2 == con.size() && c1 == 2))
        a[con.size()]++;
    }
  for (int i = 1; i <= k; i++)
    if (a[i] > 0) unz.push_back(i);
  f[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j : unz)
      if (i >= j) f[i] = (f[i] + 1ll * f[i - j] * a[j] % 998244353) % 998244353;
  printf("%d\n", f[m]);
  return 0;
}
