#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353, G = 3;
void ADD(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void SUB(int &x, int y) {
  x -= y;
  if (x < 0) x += mod;
}
long long power(long long a, int b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ans;
}
namespace PolyOp {
int t1[1 << 20], t2[1 << 20], num[1 << 20];
long long W[1 << 20 | 10];
int LSTN;
long long invn;
void PreCalcNTT(int N) {
  if (N == LSTN) return;
  LSTN = N, invn = power(N, mod - 2);
  for (int i = 0; i < N; i++) {
    num[i] = num[i >> 1] >> 1 | (i & 1 ? N >> 1 : 0);
  }
  W[0] = 1, W[1] = power(G, (mod - 1) / N);
  for (int i = 2; i <= N; i++) W[i] = W[i - 1] * W[1] % mod;
}
void NTT(int *f, int N, int flag) {
  for (int i = 0; i < N; i++) {
    if (i < num[i]) swap(f[i], f[num[i]]);
  }
  for (int len = 2; len <= N; len <<= 1) {
    for (int i = 0; i < N; i += len) {
      int *fl = f + i, *fr = fl + len / 2, step = flag ? -N / len : N / len;
      long long *w = flag ? W + N : W;
      for (int j = 0; j < len / 2; j++) {
        int tmp = *fr * *w % mod;
        SUB(*fr = *fl, tmp), ADD(*fl, tmp);
        fl++, fr++, w += step;
      }
    }
  }
  if (flag) {
    for (int i = 0; i < N; i++) {
      f[i] = f[i] * invn % mod;
    }
  }
}
void polyinv(int *f, int n, int *g) {
  if (n == 1) {
    g[0] = power(f[0], mod - 2);
    return;
  }
  int mid = n + 1 >> 1;
  polyinv(f, mid, g);
  int N = 1;
  while (N < mid + mid + n - 2) N <<= 1;
  PreCalcNTT(N);
  for (int i = 0; i < N; i++) {
    t1[i] = i < mid ? g[i] : 0, t2[i] = i < n ? f[i] : 0;
  }
  NTT(t1, N, 0), NTT(t2, N, 0);
  for (int i = 0; i < N; i++) {
    t1[i] = 1ll * t1[i] * t1[i] % mod * t2[i] % mod;
  }
  NTT(t1, N, 1);
  for (int i = 0; i < n; i++) {
    g[i] = (2ll * (i < mid ? g[i] : 0) - t1[i] + mod) % mod;
  }
}
}  // namespace PolyOp
int n, m, k;
vector<vector<int> > a;
int cnt[300010];
int f[300010], g[300010];
int suc[300010], pre[300010];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    int k;
    scanf("%d", &k);
    int lst = -1;
    while (k--) {
      int x;
      scanf("%d", &x);
      if (lst != -1) {
        if (suc[lst] == 0)
          suc[lst] = x;
        else if (suc[lst] != x)
          suc[lst] = -1;
        if (pre[x] == 0)
          pre[x] = lst;
        else if (pre[x] != lst)
          pre[x] = -1;
      }
      lst = x;
    }
  }
  f[0] = 1;
  for (int i = 1; i <= k; i++) {
    if (pre[i] == 0) {
      int cur = i, cnt = 1;
      while (1) {
        if (suc[cur] == -1) goto BAD;
        if (suc[cur] == 0) break;
        cur = suc[cur], cnt++;
        if (pre[cur] == -1) goto BAD;
      }
      f[cnt]--;
    BAD:;
    }
  }
  for (int i = 1; i <= m; i++) {
    f[i] = (f[i] + mod) % mod;
  }
  PolyOp ::polyinv(f, m + 1, g);
  printf("%d\n", g[m]);
  return 0;
}

### Prompt

Please formulate 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, G = 3;
void ADD(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void SUB(int &x, int y) {
  x -= y;
  if (x < 0) x += mod;
}
long long power(long long a, int b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ans;
}
namespace PolyOp {
int t1[1 << 20], t2[1 << 20], num[1 << 20];
long long W[1 << 20 | 10];
int LSTN;
long long invn;
void PreCalcNTT(int N) {
  if (N == LSTN) return;
  LSTN = N, invn = power(N, mod - 2);
  for (int i = 0; i < N; i++) {
    num[i] = num[i >> 1] >> 1 | (i & 1 ? N >> 1 : 0);
  }
  W[0] = 1, W[1] = power(G, (mod - 1) / N);
  for (int i = 2; i <= N; i++) W[i] = W[i - 1] * W[1] % mod;
}
void NTT(int *f, int N, int flag) {
  for (int i = 0; i < N; i++) {
    if (i < num[i]) swap(f[i], f[num[i]]);
  }
  for (int len = 2; len <= N; len <<= 1) {
    for (int i = 0; i < N; i += len) {
      int *fl = f + i, *fr = fl + len / 2, step = flag ? -N / len : N / len;
      long long *w = flag ? W + N : W;
      for (int j = 0; j < len / 2; j++) {
        int tmp = *fr * *w % mod;
        SUB(*fr = *fl, tmp), ADD(*fl, tmp);
        fl++, fr++, w += step;
      }
    }
  }
  if (flag) {
    for (int i = 0; i < N; i++) {
      f[i] = f[i] * invn % mod;
    }
  }
}
void polyinv(int *f, int n, int *g) {
  if (n == 1) {
    g[0] = power(f[0], mod - 2);
    return;
  }
  int mid = n + 1 >> 1;
  polyinv(f, mid, g);
  int N = 1;
  while (N < mid + mid + n - 2) N <<= 1;
  PreCalcNTT(N);
  for (int i = 0; i < N; i++) {
    t1[i] = i < mid ? g[i] : 0, t2[i] = i < n ? f[i] : 0;
  }
  NTT(t1, N, 0), NTT(t2, N, 0);
  for (int i = 0; i < N; i++) {
    t1[i] = 1ll * t1[i] * t1[i] % mod * t2[i] % mod;
  }
  NTT(t1, N, 1);
  for (int i = 0; i < n; i++) {
    g[i] = (2ll * (i < mid ? g[i] : 0) - t1[i] + mod) % mod;
  }
}
}  // namespace PolyOp
int n, m, k;
vector<vector<int> > a;
int cnt[300010];
int f[300010], g[300010];
int suc[300010], pre[300010];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    int k;
    scanf("%d", &k);
    int lst = -1;
    while (k--) {
      int x;
      scanf("%d", &x);
      if (lst != -1) {
        if (suc[lst] == 0)
          suc[lst] = x;
        else if (suc[lst] != x)
          suc[lst] = -1;
        if (pre[x] == 0)
          pre[x] = lst;
        else if (pre[x] != lst)
          pre[x] = -1;
      }
      lst = x;
    }
  }
  f[0] = 1;
  for (int i = 1; i <= k; i++) {
    if (pre[i] == 0) {
      int cur = i, cnt = 1;
      while (1) {
        if (suc[cur] == -1) goto BAD;
        if (suc[cur] == 0) break;
        cur = suc[cur], cnt++;
        if (pre[cur] == -1) goto BAD;
      }
      f[cnt]--;
    BAD:;
    }
  }
  for (int i = 1; i <= m; i++) {
    f[i] = (f[i] + mod) % mod;
  }
  PolyOp ::polyinv(f, m + 1, g);
  printf("%d\n", g[m]);
  return 0;
}

```