#include <bits/stdc++.h>
using namespace std;
const long long dv = 998244353;
long long n, m, d, nn = 0, dsu[300069], cc[300069], ce[300069], dg[2][300069],
                   cdg[2][300069], fq[300069], dp[300069], ex[300069];
map<long long, long long> ve[300069];
long long fd(long long x) {
  if (dsu[x] != x) {
    dsu[x] = fd(dsu[x]);
  }
  return dsu[x];
}
int main() {
  long long i, j, ii, k, l, sz;
  scanf("%lld%lld%lld", &n, &m, &d);
  for (i = 1; i <= d; i++) {
    dsu[i] = i;
    cc[i] = 1;
  }
  for (i = 0; i < n; i++) {
    scanf("%lld", &sz);
    for (j = 0; j < sz; j++) {
      scanf("%lld", &k);
      if (j && !ve[l][k]) {
        ve[l][k] = 1;
        if (fd(k) != fd(l)) {
          cc[fd(l)] += cc[fd(k)];
          ce[fd(l)] += ce[fd(k)];
          for (ii = 0; ii < 2; ii++) {
            cdg[ii][fd(l)] += cdg[ii][fd(k)];
          }
          dsu[fd(k)] = l;
        }
        ce[fd(k)]++;
        for (ii = 0; ii < 2; ii++) {
          cdg[ii][fd(k)] -= dg[ii][k] == 1;
          dg[ii][k]++;
          cdg[ii][fd(k)] += dg[ii][k] == 1;
          swap(k, l);
        }
      }
      l = k;
    }
  }
  for (i = 1; i <= d; i++) {
    if (fd(i) == i && ce[i] == cc[i] - 1 && cdg[0][i] == cc[i] - 1 &&
        cdg[1][i] == cc[i] - 1) {
      fq[cc[i]]++;
    }
  }
  for (i = 1; i <= d; i++) {
    if (fq[i]) {
      nn++;
      ex[nn] = i;
    }
  }
  dp[0] = 1;
  for (i = 1; i <= m; i++) {
    for (j = 1; j <= nn; j++) {
      if (i >= ex[j]) {
        dp[i] = (dp[i] + dp[i - ex[j]] * fq[ex[j]]) % dv;
      }
    }
  }
  printf("%lld\n", dp[m]);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long dv = 998244353;
long long n, m, d, nn = 0, dsu[300069], cc[300069], ce[300069], dg[2][300069],
                   cdg[2][300069], fq[300069], dp[300069], ex[300069];
map<long long, long long> ve[300069];
long long fd(long long x) {
  if (dsu[x] != x) {
    dsu[x] = fd(dsu[x]);
  }
  return dsu[x];
}
int main() {
  long long i, j, ii, k, l, sz;
  scanf("%lld%lld%lld", &n, &m, &d);
  for (i = 1; i <= d; i++) {
    dsu[i] = i;
    cc[i] = 1;
  }
  for (i = 0; i < n; i++) {
    scanf("%lld", &sz);
    for (j = 0; j < sz; j++) {
      scanf("%lld", &k);
      if (j && !ve[l][k]) {
        ve[l][k] = 1;
        if (fd(k) != fd(l)) {
          cc[fd(l)] += cc[fd(k)];
          ce[fd(l)] += ce[fd(k)];
          for (ii = 0; ii < 2; ii++) {
            cdg[ii][fd(l)] += cdg[ii][fd(k)];
          }
          dsu[fd(k)] = l;
        }
        ce[fd(k)]++;
        for (ii = 0; ii < 2; ii++) {
          cdg[ii][fd(k)] -= dg[ii][k] == 1;
          dg[ii][k]++;
          cdg[ii][fd(k)] += dg[ii][k] == 1;
          swap(k, l);
        }
      }
      l = k;
    }
  }
  for (i = 1; i <= d; i++) {
    if (fd(i) == i && ce[i] == cc[i] - 1 && cdg[0][i] == cc[i] - 1 &&
        cdg[1][i] == cc[i] - 1) {
      fq[cc[i]]++;
    }
  }
  for (i = 1; i <= d; i++) {
    if (fq[i]) {
      nn++;
      ex[nn] = i;
    }
  }
  dp[0] = 1;
  for (i = 1; i <= m; i++) {
    for (j = 1; j <= nn; j++) {
      if (i >= ex[j]) {
        dp[i] = (dp[i] + dp[i - ex[j]] * fq[ex[j]]) % dv;
      }
    }
  }
  printf("%lld\n", dp[m]);
}

```