#include <bits/stdc++.h>
#include <unordered_set>
using namespace std;

#define st first
#define nd second
#define pb push_back
#define mp make_pair
#define klar(v) memset(v, 0, sizeof(v))
#define bust ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define gcd(a, b) __gcd(a, b);
#define debug cout << "chuj" << endl;
#define endl "\n"

typedef vector<int> vi;
typedef vector<pair<int, int> > vpii;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef long long ll;
const int mod = 1e8+37, mod2 = 1e9+7, prime = 137, maxn = 1e5+100;

vi arr[maxn];
int primes[maxn], primes2[maxn];
set <pii> keks[maxn];
pii gethasz(int n, int h, int it, ll hasz){
    int w = n-it;
    ll hasz2 = (1LL*mod+hasz-(1LL*arr[h][it]*primes[w])%mod)%mod;
    hasz = (1LL*mod2+hasz-(1LL*arr[h][it]*primes[w])%mod2)%mod2;
    return mp(hasz2, hasz);
}

int main(){
    primes[0] = 1;
    for(int i = 1; i < maxn; i++)
        primes[i] = (1LL*primes[i-1]*prime)%mod;
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++){
        int xn;
        scanf("%d", &xn);
        arr[i].reserve(xn+5);
        arr[i].pb(0);
        for(int j = 1; j <= xn; j++){
            int x;
            scanf("%d", &x);
            arr[i].pb(x);
//            arr[i][j]++;
        }
        ll ahasz = 0;
        for(int j = 1; j <= xn; j++)
            ahasz = (ahasz*prime+arr[i][j])%mod;
        for(int j = 1; j <= xn; j++)
            keks[xn-1].insert(gethasz(xn, i, j, ahasz));
    }
    while(m--){
        int xn;
        scanf("%d", &xn);
        vi ac;
        ac.reserve(xn+3);
        ll allhasz = 0;
        ac.pb(0);
        for(int i = 1; i <= xn; i++){
            int x;
            scanf("%d", &x);
            ac.pb(x);
//            ac[i]++;
            allhasz = (allhasz*prime+ac[i])%mod;
        }
        bool ans = 0;
        for(int i = 1; i <= xn; i++){
            int h = xn-i;
            int szuk1 = (mod+allhasz-(1LL*ac[i]*primes[h])%mod)%mod;
            int szuk2 = (mod2+allhasz-(1LL*ac[i]*primes[h])%mod2)%mod2;
            if(keks[xn-1].find(mp(szuk1, szuk2)) != keks[xn-1].end())ans = 1;
        }
        if(ans)printf("TAK\n");
        else printf("NIE\n");
    }
}