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
#include<bits/stdc++.h>
#define ls i<<1
#define rs i<<1|1
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=3E5+10;
const int inf=1<<30;
auto max(auto l,auto r){return l>r?l:r;}
auto min(auto l,auto r){return l<r?l:r;}
ll read()
{
    ll a=0;
    char b=1,c;
    do if((c=getchar())==45)b=-1;while(c<48||c>57);
    do a=(a<<3)+(a<<1)+(c&15);while((c=getchar())>47&&c<58);
    return a*b;
}
void write(ll x,char c)
{
    if(x<0)putchar(45),x=-x;
    char a[20],s=0;
    do a[++s]=x%10|48;while(x/=10);
    do putchar(a[s]);while(--s);
    putchar(c);
}
int pre[N],nex[N],a[N],f[N];
bool in[N],vis[N];
vector<pii>q;
int g(int x){return x<0?0:f[x];}
int dfs(int x)
{
    if(!x)return 0;
    if(vis[x])puts("0"),exit(0);
    vis[x]=1;
    return dfs(nex[x])+1;
}
int main()
{
    const int p=998244353;
    int n=read(),m=read(),s=read();
    while(n--)
    {
        for(int y=read(),l=read(),r;--y;l=r)
        {
            r=read();
            if(nex[l]^r)
            {
                if(nex[l]){puts("0");return 0;}
                in[nex[l]=r]=1;
            }
        }
    }
    for(int i=1;i<=s;++i)
    {
        if(!in[i])
        {
            ++a[dfs(i)];
        }
    }
    for(int i=1;i<=s;++i)
    {
        if(!vis[i])
        {
            puts("0");
            return 0;
        }
    }
    for(int i=1;i<=s;++i)
    {
        if(a[i])
        {
            q.pb({i,a[i]});
        }
    }
    f[0]=1;
    for(int i=1;i<=m;++i)
    {
        for(auto [x,y]:q)
        {
            f[i]=(f[i]+g(i-x)*1LL*y)%p;
        }
    }
    write(f[m],'\n');
}
