#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 1e6 + 5;
const long long MOD = 998244353;
const long long MOD2 = 2010405347;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 315;
inline long long pw(long long x, long long y, long long md = MOD) {
  long long ret = 1;
  x %= md;
  while (y) {
    if (y & 1) ret = ret * x % md;
    x = x * x % md, y >>= 1;
  }
  return ret;
}
inline void add(int &a, int b, int md = MOD) {
  a = a + b >= md ? a + b - md : a + b;
}
int n, m, k, le[MAX], ri[MAX], x[MAX], c, cnt[MAX], nx[MAX], ls[MAX], hit, nw,
    p[MAX], a, b;
bool st[MAX], tmp;
long long ans, dp[MAX];
vector<pair<int, int> > v;
long long f(int rem) {
  if (!rem) return 1;
  if (dp[rem] != -1) return dp[rem];
  long long ret = 0;
  for (pair<int, int> &i : v) {
    if (i.first > rem) break;
    ret += f(rem - i.first) * i.second;
  }
  return dp[rem] = ret % MOD;
}
int par(int z) {
  if (p[z] == z) return z;
  return p[z] = par(p[z]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> c;
  for (int i = (1); i <= (c); ++i) p[i] = i, st[i] = 1;
  for (int i = (1); i <= (n); ++i) {
    cin >> k;
    for (int i = (1); i <= (k); ++i) cin >> x[i];
    tmp = 1;
    for (int i = (2); i <= (k); ++i) {
      if ((nx[x[i - 1]] && nx[x[i - 1]] != x[i]) ||
          (ls[x[i]] && ls[x[i]] != x[i - 1]) || x[i] == x[i - 1])
        tmp = 0;
      a = par(x[i - 1]), b = par(x[i]);
      tmp &= st[a] && st[b];
      p[b] = a;
      st[a] = tmp;
      if (!tmp) continue;
      nx[x[i - 1]] = x[i];
      ls[x[i]] = x[i - 1];
    }
  }
  for (int i = (1); i <= (c); ++i)
    if (!ls[i] && st[par(i)]) {
      if (!nx[i])
        ++cnt[1];
      else {
        nw = i, hit = 1;
        while (nx[nw]) nw = nx[nw], ++hit;
        ++cnt[hit];
      }
    }
  for (int i = (1); i <= (m); ++i)
    if (cnt[i]) v.push_back({i, cnt[i]});
  memset(&dp, -1, sizeof dp);
  cout << f(m) << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#pragma GCC optimize("O3")
#pragma GCC target("tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 1e6 + 5;
const long long MOD = 998244353;
const long long MOD2 = 2010405347;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 315;
inline long long pw(long long x, long long y, long long md = MOD) {
  long long ret = 1;
  x %= md;
  while (y) {
    if (y & 1) ret = ret * x % md;
    x = x * x % md, y >>= 1;
  }
  return ret;
}
inline void add(int &a, int b, int md = MOD) {
  a = a + b >= md ? a + b - md : a + b;
}
int n, m, k, le[MAX], ri[MAX], x[MAX], c, cnt[MAX], nx[MAX], ls[MAX], hit, nw,
    p[MAX], a, b;
bool st[MAX], tmp;
long long ans, dp[MAX];
vector<pair<int, int> > v;
long long f(int rem) {
  if (!rem) return 1;
  if (dp[rem] != -1) return dp[rem];
  long long ret = 0;
  for (pair<int, int> &i : v) {
    if (i.first > rem) break;
    ret += f(rem - i.first) * i.second;
  }
  return dp[rem] = ret % MOD;
}
int par(int z) {
  if (p[z] == z) return z;
  return p[z] = par(p[z]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> c;
  for (int i = (1); i <= (c); ++i) p[i] = i, st[i] = 1;
  for (int i = (1); i <= (n); ++i) {
    cin >> k;
    for (int i = (1); i <= (k); ++i) cin >> x[i];
    tmp = 1;
    for (int i = (2); i <= (k); ++i) {
      if ((nx[x[i - 1]] && nx[x[i - 1]] != x[i]) ||
          (ls[x[i]] && ls[x[i]] != x[i - 1]) || x[i] == x[i - 1])
        tmp = 0;
      a = par(x[i - 1]), b = par(x[i]);
      tmp &= st[a] && st[b];
      p[b] = a;
      st[a] = tmp;
      if (!tmp) continue;
      nx[x[i - 1]] = x[i];
      ls[x[i]] = x[i - 1];
    }
  }
  for (int i = (1); i <= (c); ++i)
    if (!ls[i] && st[par(i)]) {
      if (!nx[i])
        ++cnt[1];
      else {
        nw = i, hit = 1;
        while (nx[nw]) nw = nx[nw], ++hit;
        ++cnt[hit];
      }
    }
  for (int i = (1); i <= (m); ++i)
    if (cnt[i]) v.push_back({i, cnt[i]});
  memset(&dp, -1, sizeof dp);
  cout << f(m) << '\n';
  return 0;
}

```