#include<bits/stdc++.h>
#include<algorithm>
#include<random>
#include<chrono>
#include<cstdlib>
#include<ctime>
#include<numeric>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<queue>
#include<iomanip>
#define int long long
#define ll long long
#define L LLONG_MAX
#define fi first
#define se second
#define pii pair<int,int>
#define sz(a) ((int)a.size())
#define FOR(i,j,k) for(int i=j;i<=k;i++)
#define REP(i,k,j) for(int i=k;i>=j;i--)
#define FORD(i,a) for(auto i:a)
using namespace std;
const int NMAX = 500000 + 5;
int n, q;
int c[NMAX];
vector<int> arr[NMAX]; // keys in each room (like a[i])
vector<int> pos[NMAX]; // positions (rooms) that contain key type t
int lef[NMAX], rig[NMAX];
int X[NMAX], Y[NMAX];
bool inside(int l, int r, int key){
if (key <= 0 || key >= NMAX) return false;
auto &v = pos[key];
if (v.empty()) return false;
auto it = lower_bound(v.begin(), v.end(), l);
return it != v.end() && *it <= r;
}
void input(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
if (!(cin >> n)) exit(0);
FOR(i,1,n-1) cin >> c[i];
FOR(i,1,n){
int b; cin >> b;
arr[i].resize(b);
FOR(j,0,b-1){
cin >> arr[i][j];
// push room index i into pos[key], we iterate i increasing so pos[key] stays sorted
if (arr[i][j] > 0 && arr[i][j] < NMAX) pos[ arr[i][j] ].push_back(i);
}
}
cin >> q;
FOR(i,1,q) cin >> X[i] >> Y[i];
}
void solve(){
// compute lef/rig for each starting room i
FOR(i,1,n){
lef[i] = rig[i] = i;
bool nxt = true;
while (nxt){
nxt = false;
// try expand left using merge with precomputed block if corridor c[lef[i]-1] is present in current [lef,rig]
if (lef[i] > 1 && inside(lef[i], rig[i], c[lef[i]-1])){
// merge with block of lef[i]-1 (which has been computed because lef[i]-1 < i)
rig[i] = max(rig[i], rig[lef[i]-1]);
lef[i] = lef[lef[i]-1];
nxt = true;
continue; // after merging left, try again from top
}
// else try expand right by checking corridor c[rig[i]]
if (rig[i] < n && inside(lef[i], rig[i], c[rig[i]])){
rig[i]++; // include the next room on right
nxt = true;
// note: after rig[i]++ we don't immediately merge with precomputed blocks to keep same logic as original;
// on next outer loop iteration we may merge via lef[rig[i]] if applicable.
}
}
}
// answer queries
FOR(i,1,q){
int x = X[i], y = Y[i];
bool ok = (lef[x] <= y && y <= rig[x]);
cout << (ok ? "YES" : "NO") << '\n';
}
}
signed main(){
input();
solve();
return 0;
}
