#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 500005;
int N, C[MX], L[MX], R[MX];
int needL[MX], needR[MX];
vi posi[MX], Rneed[MX], Lneed[MX];
priority_queue<int> p[MX];
set<int> s;
deque<pi> d;
void tri0(int i) {
auto it = s.lb(needL[i]); if (it == s.begin()) return;
int j = *prev(it); if (i > j) return;
if (i == 1) { d.pb({i,N}); return; }
int num = j-i+1; int lef = num;
while (lef) {
int LEF = min(d.front().s,lef);
d.front().s -= LEF, lef -= LEF;
if (d.front().s == 0) d.pop_front();
}
d.push_front({i,num});
}
void tri1(int i) {
auto it = s.ub(needR[i]); if (it == s.end()) return;
int j = *it; if (j > i) return;
if (i == N) { d.pb({i,N}); return; }
int num = i-j+1; int lef = num;
while (lef) {
int LEF = min(d.front().s,lef);
d.front().s -= LEF, lef -= LEF;
if (d.front().s == 0) d.pop_front();
}
d.push_front({i,num});
}
void del() {
d.front().s --;
if (!d.front().s) d.pop_front();
}
void gen() {
FOR(i,1,N+1) for (int j: posi[i]) p[j].push(-i);
FOR(i,1,N+1) {
needL[i] = (sz(p[C[i-1]]) ? -p[C[i-1]].top() : N+1);
Lneed[needL[i]].pb(i);
for (int j: posi[i]) p[j].pop();
}
FOR(i,1,N+1) for (int j: posi[i]) p[j].push(i);
FORd(i,1,N+1) {
needR[i] = (sz(p[C[i]]) ? p[C[i]].top() : 0);
Rneed[needR[i]].pb(i);
for (int j: posi[i]) p[j].pop();
}
FOR(i,1,N+1) {
for (int j: Rneed[i-1]) s.insert(j);
tri0(i);
L[i] = d.front().f;
del();
}
s.clear(); assert(!sz(d));
FORd(i,1,N+1) {
for (int j: Lneed[i+1]) s.insert(j);
tri1(i);
R[i] = d.front().f;
del();
}
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N; FOR(i,1,N) cin >> C[i];
FOR(i,1,N+1) {
int B; cin >> B;
F0R(j,B) {
int A; cin >> A;
posi[i].pb(A);
}
}
gen();
int Q; cin >> Q;
F0R(i,Q) {
int X,Y; cin >> X >> Y;
if (L[X] <= Y && Y <= R[X]) cout << "YES";
else cout << "NO";
cout << "\n";
}
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/
