#include <bits/stdc++.h>
using namespace std;
const long long N = 3 * 1e5 + 100;
long long n, m, k, vi[N], l[N], r[N], rev[N << 2], len;
long long dp[N], g[N], pr[N];
long long w, b[N], c[N], Inv[N];
pair<long long, long long> t[N];
vector<long long> A[N], p[N], f[25];
inline void add(long long &a, long long b) {
  a = ((a + b >= 998244353) ? a + b - 998244353 : a + b);
}
inline void del(long long &a, long long b) {
  a = ((a - b < 0) ? a - b + 998244353 : a - b);
}
inline void mul(long long &a, long long b) { a = a * b % 998244353; }
inline long long m_pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) mul(ans, a);
    b >>= 1;
    mul(a, a);
  }
  return ans;
}
inline long long read() {
  long long f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
void prework() {
  long long d = 20;
  for (long long i = 1; i <= d; i++) {
    f[i].resize(1 << i);
    f[i][0] = 1;
    f[i][1] = m_pow(3, (998244353 - 1) >> i);
    for (long long j = 2; j < (1 << i); j++)
      f[i][j] = f[i][j - 1] * f[i][1] % 998244353;
  }
}
inline void change(long long len) {
  for (long long i = 0; i < len; i++) {
    rev[i] = rev[i >> 1] >> 1;
    if (i & 1) rev[i] |= (len >> 1);
  }
}
inline void DFT(vector<long long> &y, long long len, long long v) {
  for (long long i = 0; i < len; i++)
    if (i < rev[i]) swap(y[i], y[rev[i]]);
  for (long long i = 1, d = 1; i < len; i <<= 1, d++) {
    for (long long j = 0; j < len; j += (i << 1)) {
      long long *W = &f[d][0], a, b;
      for (long long k = j; k < j + i; k++) {
        a = y[k], b = y[k + i] * (*W++) % 998244353;
        y[k] = y[k + i] = a;
        add(y[k], b);
        del(y[k + i], b);
      }
    }
  }
  if (v == -1) {
    reverse(y.begin() + 1, y.end());
    long long invlen = m_pow(len, 998244353 - 2);
    for (long long i = 0; i < len; i++) mul(y[i], invlen);
  }
}
vector<long long> operator*(vector<long long> a, vector<long long> b) {
  if (a.empty()) return b;
  if (b.empty()) return a;
  long long n = (long long)a.size(), m = (long long)b.size();
  len = 1;
  while (len < n + m) len <<= 1;
  change(len);
  a.resize(len);
  b.resize(len);
  DFT(a, len, 1);
  DFT(b, len, 1);
  for (long long i = 0; i < len; i++) mul(a[i], b[i]);
  DFT(a, len, -1);
  a.resize(n + m - 1);
  return a;
}
void solve(long long l, long long r) {
  if (l == r) {
    if (l == 0) dp[l] = 1;
    return;
  }
  long long mid = (l + r) >> 1;
  solve(l, mid);
  vector<long long> A;
  A = vector<long long>(dp + l, dp + mid + 1) *
      vector<long long>(g, g + r - l + 1);
  for (long long i = mid + 1; i <= r; i++) add(dp[i], A[i - l]);
  solve(mid + 1, r);
}
void del(long long id) {
  vi[id] = 1;
  for (long long i : A[id])
    if (t[i].first != -1) {
      vi[t[i].first] = 1;
      t[i] = make_pair(-1, -1);
    }
}
void check_twice(long long id) {
  static long long cnt[N];
  for (long long i : A[id]) cnt[i]++;
  bool bl = 1;
  for (long long i : A[id])
    if (cnt[i] > 1) bl = 0;
  if (!bl) del(id);
  for (long long i : A[id]) cnt[i] = 0;
}
void upd(long long x) {
  if (t[x].first <= 0) return;
  if (vi[t[x].first]) t[x] = make_pair(-1, -1);
}
void merge(long long &x, long long &y) {
  if ((long long)p[x].size() < (long long)p[y].size()) {
    l[y] = l[y] - (long long)p[x].size();
    for (long long i : p[x]) t[i] = make_pair(y, t[i].second - l[x] + l[y]);
    p[y].insert(p[y].end(), p[x].begin(), p[x].end());
    vi[x] = 1;
    x = y;
  } else {
    r[x] = r[x] + (long long)p[y].size();
    for (long long i : p[y]) t[i] = make_pair(x, r[x] - (r[y] - t[i].second));
    p[x].insert(p[x].begin(), p[y].begin(), p[y].end());
    vi[y] = 1;
    y = x;
  }
}
bool cmp(long long a, long long b) { return t[a].second < t[b].second; }
bool cmp1(long long a, long long b) {
  return (long long)A[a].size() > (long long)A[b].size();
}
signed main() {
  prework();
  n = read();
  m = read();
  k = read();
  for (long long i = 1; i <= n; i++) {
    long long c = read();
    A[i].resize(c);
    for (long long &j : A[i]) j = read();
    check_twice(i);
  }
  for (long long i = 1; i <= n; i++) pr[i] = i;
  sort(pr + 1, pr + 1 + n, cmp1);
  for (long long s = 1; s <= n; s++) {
    long long i = pr[s];
    if (vi[i]) continue;
    bool bl = 1;
    for (long long j : A[i]) {
      upd(j);
      if (t[j].first == -1) bl = 0;
    }
    if (!bl) {
      del(i);
      continue;
    }
    w = 0;
    for (long long j = 0; j < (long long)A[i].size();) {
      long long k = j;
      while (k < (long long)A[i].size() && t[A[i][j]].first == t[A[i][k]].first)
        k++;
      b[++w] = k - j;
      c[w] = t[A[i][j]].first;
      j = k;
    }
    bl = 1;
    for (long long j = 1, s = 0; j <= w; s += b[j++])
      if (c[j] != 0) {
        for (long long k = s; k < s + b[j] - 1; k++)
          if (t[A[i][k]].second + 1 != t[A[i][k + 1]].second) bl = 0;
        long long x = A[i][s], y = A[i][s + b[j] - 1];
        if (t[x].second != l[c[j]] && t[y].second != r[c[j]] && w > 1) bl = 0;
        if (t[x].second == l[c[j]] && j != w) bl = 0;
        if (t[y].second == r[c[j]] && j != 1) bl = 0;
      }
    if (w > 3 || (w == 3 && c[2] != 0) || !bl) {
      del(i);
      continue;
    }
    for (long long j = 1, s = 0; j <= w; s += b[j++])
      if (c[j] == 0) {
        for (long long k = s; k < s + b[j]; k++)
          t[A[i][k]] = make_pair(i, r[i]++), p[i].push_back(A[i][k]);
        c[j] = i;
      }
    r[i]--;
    for (long long i = 1; i < w; i++) merge(c[i], c[i + 1]);
  }
  for (long long i = 1; i <= n; i++)
    if (!vi[i] && !p[i].empty()) {
      g[(long long)p[i].size()]++;
    }
  long long ret = 0;
  for (long long i = 1; i <= k; i++)
    if (t[i].first == 0) ret++;
  g[1] += ret;
  solve(0, m);
  printf("%lld\n", dp[m]);
}

### Prompt

Create a solution in Cpp for 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 long long N = 3 * 1e5 + 100;
long long n, m, k, vi[N], l[N], r[N], rev[N << 2], len;
long long dp[N], g[N], pr[N];
long long w, b[N], c[N], Inv[N];
pair<long long, long long> t[N];
vector<long long> A[N], p[N], f[25];
inline void add(long long &a, long long b) {
  a = ((a + b >= 998244353) ? a + b - 998244353 : a + b);
}
inline void del(long long &a, long long b) {
  a = ((a - b < 0) ? a - b + 998244353 : a - b);
}
inline void mul(long long &a, long long b) { a = a * b % 998244353; }
inline long long m_pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) mul(ans, a);
    b >>= 1;
    mul(a, a);
  }
  return ans;
}
inline long long read() {
  long long f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
void prework() {
  long long d = 20;
  for (long long i = 1; i <= d; i++) {
    f[i].resize(1 << i);
    f[i][0] = 1;
    f[i][1] = m_pow(3, (998244353 - 1) >> i);
    for (long long j = 2; j < (1 << i); j++)
      f[i][j] = f[i][j - 1] * f[i][1] % 998244353;
  }
}
inline void change(long long len) {
  for (long long i = 0; i < len; i++) {
    rev[i] = rev[i >> 1] >> 1;
    if (i & 1) rev[i] |= (len >> 1);
  }
}
inline void DFT(vector<long long> &y, long long len, long long v) {
  for (long long i = 0; i < len; i++)
    if (i < rev[i]) swap(y[i], y[rev[i]]);
  for (long long i = 1, d = 1; i < len; i <<= 1, d++) {
    for (long long j = 0; j < len; j += (i << 1)) {
      long long *W = &f[d][0], a, b;
      for (long long k = j; k < j + i; k++) {
        a = y[k], b = y[k + i] * (*W++) % 998244353;
        y[k] = y[k + i] = a;
        add(y[k], b);
        del(y[k + i], b);
      }
    }
  }
  if (v == -1) {
    reverse(y.begin() + 1, y.end());
    long long invlen = m_pow(len, 998244353 - 2);
    for (long long i = 0; i < len; i++) mul(y[i], invlen);
  }
}
vector<long long> operator*(vector<long long> a, vector<long long> b) {
  if (a.empty()) return b;
  if (b.empty()) return a;
  long long n = (long long)a.size(), m = (long long)b.size();
  len = 1;
  while (len < n + m) len <<= 1;
  change(len);
  a.resize(len);
  b.resize(len);
  DFT(a, len, 1);
  DFT(b, len, 1);
  for (long long i = 0; i < len; i++) mul(a[i], b[i]);
  DFT(a, len, -1);
  a.resize(n + m - 1);
  return a;
}
void solve(long long l, long long r) {
  if (l == r) {
    if (l == 0) dp[l] = 1;
    return;
  }
  long long mid = (l + r) >> 1;
  solve(l, mid);
  vector<long long> A;
  A = vector<long long>(dp + l, dp + mid + 1) *
      vector<long long>(g, g + r - l + 1);
  for (long long i = mid + 1; i <= r; i++) add(dp[i], A[i - l]);
  solve(mid + 1, r);
}
void del(long long id) {
  vi[id] = 1;
  for (long long i : A[id])
    if (t[i].first != -1) {
      vi[t[i].first] = 1;
      t[i] = make_pair(-1, -1);
    }
}
void check_twice(long long id) {
  static long long cnt[N];
  for (long long i : A[id]) cnt[i]++;
  bool bl = 1;
  for (long long i : A[id])
    if (cnt[i] > 1) bl = 0;
  if (!bl) del(id);
  for (long long i : A[id]) cnt[i] = 0;
}
void upd(long long x) {
  if (t[x].first <= 0) return;
  if (vi[t[x].first]) t[x] = make_pair(-1, -1);
}
void merge(long long &x, long long &y) {
  if ((long long)p[x].size() < (long long)p[y].size()) {
    l[y] = l[y] - (long long)p[x].size();
    for (long long i : p[x]) t[i] = make_pair(y, t[i].second - l[x] + l[y]);
    p[y].insert(p[y].end(), p[x].begin(), p[x].end());
    vi[x] = 1;
    x = y;
  } else {
    r[x] = r[x] + (long long)p[y].size();
    for (long long i : p[y]) t[i] = make_pair(x, r[x] - (r[y] - t[i].second));
    p[x].insert(p[x].begin(), p[y].begin(), p[y].end());
    vi[y] = 1;
    y = x;
  }
}
bool cmp(long long a, long long b) { return t[a].second < t[b].second; }
bool cmp1(long long a, long long b) {
  return (long long)A[a].size() > (long long)A[b].size();
}
signed main() {
  prework();
  n = read();
  m = read();
  k = read();
  for (long long i = 1; i <= n; i++) {
    long long c = read();
    A[i].resize(c);
    for (long long &j : A[i]) j = read();
    check_twice(i);
  }
  for (long long i = 1; i <= n; i++) pr[i] = i;
  sort(pr + 1, pr + 1 + n, cmp1);
  for (long long s = 1; s <= n; s++) {
    long long i = pr[s];
    if (vi[i]) continue;
    bool bl = 1;
    for (long long j : A[i]) {
      upd(j);
      if (t[j].first == -1) bl = 0;
    }
    if (!bl) {
      del(i);
      continue;
    }
    w = 0;
    for (long long j = 0; j < (long long)A[i].size();) {
      long long k = j;
      while (k < (long long)A[i].size() && t[A[i][j]].first == t[A[i][k]].first)
        k++;
      b[++w] = k - j;
      c[w] = t[A[i][j]].first;
      j = k;
    }
    bl = 1;
    for (long long j = 1, s = 0; j <= w; s += b[j++])
      if (c[j] != 0) {
        for (long long k = s; k < s + b[j] - 1; k++)
          if (t[A[i][k]].second + 1 != t[A[i][k + 1]].second) bl = 0;
        long long x = A[i][s], y = A[i][s + b[j] - 1];
        if (t[x].second != l[c[j]] && t[y].second != r[c[j]] && w > 1) bl = 0;
        if (t[x].second == l[c[j]] && j != w) bl = 0;
        if (t[y].second == r[c[j]] && j != 1) bl = 0;
      }
    if (w > 3 || (w == 3 && c[2] != 0) || !bl) {
      del(i);
      continue;
    }
    for (long long j = 1, s = 0; j <= w; s += b[j++])
      if (c[j] == 0) {
        for (long long k = s; k < s + b[j]; k++)
          t[A[i][k]] = make_pair(i, r[i]++), p[i].push_back(A[i][k]);
        c[j] = i;
      }
    r[i]--;
    for (long long i = 1; i < w; i++) merge(c[i], c[i + 1]);
  }
  for (long long i = 1; i <= n; i++)
    if (!vi[i] && !p[i].empty()) {
      g[(long long)p[i].size()]++;
    }
  long long ret = 0;
  for (long long i = 1; i <= k; i++)
    if (t[i].first == 0) ret++;
  g[1] += ret;
  solve(0, m);
  printf("%lld\n", dp[m]);
}

```