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

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};
    }
}

//{l, r} に対して、{lmax, rmin} を返す
pair<ll, ll> op(pair<ll, ll> a, pair<ll, ll> b) {
    return {max(a.first, b.first), min(a.second, b.second)};
}

pair<ll, ll> e() {
    const ll INF = 4e18;
    return {-INF, INF};
}

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);
    for (int k = 0; k < 3; k++) {
        vector<pair<ll, ll>> pq(N);
        for (int i = 0; i < N; i++) {
            pq[i] = convert(A[i], B[i], D[i], k);
        } 
        atcoder::segtree<pair<ll, ll>, op, e> seg(pq);
        for (int i = 0; i < Q; i++) {
            auto[resp, resq] = seg.prod(L[i]-1, R[i]);
            auto[sp, sq] = pq[S[i]-1];
            if (resp < sq && resq > sp) ; //ok
            else ans[i] = false; //ng
        }
    }
    for (int i = 0; i < Q; i++) {
        cout << (ans[i]? "Yes" : "No") << endl;
    }
}