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

using ll = long long;

pair<ll, ll> convert(ll a, ll b, ll d, int type) {
    if (type == 0) {
        //__が分離直線, |が分離軸
        return {b-d, b};
    }
    if (type == 1) {
        // ／が分離直線, ＼が分離軸
        return {a-b, a-b+2*d};
    }
    if (type == 2) {
        // ＼が分離直線, ／が分離軸
        return {a+b-2*d, a+b};
    }
}

const ll INF = 8e18;
ll opmin(ll a, ll b) {return min(a, b);}
ll emin() {return INF;}
ll opmax(ll a, ll b) {return max(a, b);}
ll emax() {return -INF;}

int main() {
    //入力
    int N; cin >> N;
    vector<ll> A(N), B(N), D(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i] >> B[i] >> D[i];
    }
    int Q; cin >> Q;
    vector<int> S(Q), L(Q), R(Q);
    for (int i = 0; i < Q; i++) {
        cin >> S[i] >> L[i] >> R[i];
    }
    //答え
    vector<bool> ans(Q, true);
    //各分離直線の向きについて調べる
    for (int k = 0; k < 3; k++) {
    	vector<ll> p(N), q(N);
        //投影する
        for (int i = 0; i < N; i++) {
        	auto[tp, tq] = convert(A[i], B[i], D[i], k);
        	p[i] = tp; q[i] = tq;
        } 
        segtree<ll, opmax, emax> pseg(p);
        segtree<ll, opmin, emin> qseg(q);
        for (int i = 0; i < Q; i++) {
        	ll resp = pseg.prod(L[i]-1, R[i]), resq = qseg.prod(L[i]-1, R[i]);
        	ll sp = p[S[i]-1], sq = q[S[i]-1];
            if (resp < sq and resq > sp) ; //ok
            else ans[i] = false; //ng
        }
    }
    for (int i = 0; i < Q; i++) {
        cout << (ans[i]? "Yes" : "No") << endl;
    }
}