#888

//A
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10,mod=1e9+7;
string s;
int n,t,a[N],f[N],res,num,ans,m,k,p;
bool vis[N];
signed main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n>>m>>k>>p;
        res=0;
        for(int i=0;i<n;i++){
            cin>>num;
            if(abs(num-p)%k!=0) continue;
            if(abs(num-p)>m*k) continue;
            res++;
        }
        cout<<res<<endl;
    }
    return 0;
}

//B
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10,mod=1e9+7;
string s;
int n,t,a[N],f[N],res,num,ans,m;
bool vis[N];
signed main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n;
        for(int i=0;i<n;i++) cin>>a[i],f[i]=a[i];
        sort(f,f+n);
        for(int i=0;i<n;i++){
            if(f[i]%2==0&&a[i]%2==0) continue;
            else if(f[i]%2!=0&&a[i]%2!=0) continue;
            else{
                cout<<"NO"<<endl;
                goto nexts;
            }
        }
        cout<<"YES"<<endl;
        nexts:;
    }
    return 0;
}

//c
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10,mod=1e9+7;
string s;
int n,t,a[N],f[N],res,num,ans,m,k;
bool vis[N];
signed main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n>>k;
        int num1=0,num2=0;
        for(int i=0;i<n;i++) cin>>a[i];
        for(int i=0;i<n;i++){
            if(a[i]==a[0]) num1++;
            if(num1>=k&&a[i]==a[n-1]) num2++;
        }
        if((num1>=k&&num2>=k)||(a[0]==a[n-1]&&num1>=k)) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}

//d
//可以观察规律,当差分数组有两个数未出现,则第一个数组的合一定是序列的前缀和
//如果只有一个数没出现,则用序列的前缀和减去第一个数组的合就是那个剩下的数
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10,mod=1e9+7;
string s;
int n,t,a[N],f[N],res,num,ans,m,k,c[N];
map<int,int>vis;
void solve()
{
    vis.clear();
    cin>>n;
    int num=0,ans=0,sum=(n*(n+1))/2,flag=1;    
    for(int i=1;i<n;i++){
        cin>>a[i];
        c[i]=a[i]-a[i-1];
        vis[c[i]]=1,num+=c[i];
    }
    for(int i=1;i<=n;i++) if(!vis[i]) ans++;
    if((ans==2&&num==sum)||(ans==1&&((sum-num)>=1&&(sum-num)<=n))) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}
signed main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

//d
//记忆化搜索
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+10,mod=1e9+7;
string s;
int n,t,a[N],f[N],res,num,ans,m,k;
int e[N*2],ne[N*2],h[N*2],idx;
bool vis[N];
void add(int a,int b){
    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
void dfs(int x)
{
     if(vis[x]) return;
     vis[x]=true;
     int ans=0,cnt=0;
     for(int i=h[x];~i;i=ne[i]){
        int j=e[i];
        dfs(j);
        ans+=a[j],cnt++;
     }
     if(cnt) a[x]=min(a[x],ans);
}
void solve()
{
    scanf("%lld%lld", &n, &m);
    memset(vis,false,sizeof vis);
    memset(h, -1, sizeof h);
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=1;i<=m;i++) cin>>k,a[k]=0;
    for(int i=1;i<=n;i++){
        int x,y;
        scanf("%lld", &x);
       while(x--) scanf("%lld", &y),add(i,y);
    }
    for(int i=1;i<=n;i++){
        dfs(i);
        printf("%lld ",a[i]);
    }
    printf("\n");
}
signed main()
{
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}