#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &f) {
  f = 0;
  T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + 48);
  else
    print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
  print(x);
  putchar(t);
}
const int md = 998244353;
inline int add(int x, int y) {
  if (x + y >= md) return x + y - md;
  return x + y;
}
inline void addx(int &x, int y) {
  x += y;
  if (x >= md) x -= md;
}
inline int sub(int x, int y) {
  if (x < y) return x - y + md;
  return x - y;
}
inline void subx(int &x, int y) {
  x -= y;
  if (x < 0) x += md;
}
inline int mul(int x, int y) { return 1ull * x * y % md; }
inline int fpow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    y >>= 1;
    x = mul(x, x);
  }
  return ans;
}
const int 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() {
  read(n);
  read(m);
  read(k);
  for (int i = 1; i <= n; i++) {
    int len;
    read(len);
    flag = 0;
    for (int j = 1; j <= len; j++) {
      read(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] = add(dp[i], mul(dp[i - g[j].first], g[j].second));
    }
  }
  print(dp[m], '\n');
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T>
inline void read(T &f) {
  f = 0;
  T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + 48);
  else
    print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
  print(x);
  putchar(t);
}
const int md = 998244353;
inline int add(int x, int y) {
  if (x + y >= md) return x + y - md;
  return x + y;
}
inline void addx(int &x, int y) {
  x += y;
  if (x >= md) x -= md;
}
inline int sub(int x, int y) {
  if (x < y) return x - y + md;
  return x - y;
}
inline void subx(int &x, int y) {
  x -= y;
  if (x < 0) x += md;
}
inline int mul(int x, int y) { return 1ull * x * y % md; }
inline int fpow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    y >>= 1;
    x = mul(x, x);
  }
  return ans;
}
const int 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() {
  read(n);
  read(m);
  read(k);
  for (int i = 1; i <= n; i++) {
    int len;
    read(len);
    flag = 0;
    for (int j = 1; j <= len; j++) {
      read(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] = add(dp[i], mul(dp[i - g[j].first], g[j].second));
    }
  }
  print(dp[m], '\n');
  return 0;
}

```