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// Marcin Knapik Potyczki Algorytmiczne, dzień 2, Zadanie "Herbata"[B]
// złożoność O(n log n)

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

#define rall(s) s.rbegin(), s.rend()
#define f first
#define s second
#define ll long long
#define pb push_back

int q, n;

void solve(){
	cin >> q;
	while(q--){
		cin >> n;
		vector<pair<ll, ll> > jest;
		vector<pair<ll, ll> > ma_byc; 
		ll suma = 0;
		for(int i = 0; i < n; i++){
			int a, b, c;
			cin >> a >> b >> c;
			suma += a;

			jest.pb({b, a});
			ma_byc.pb({c, a});
		}

		sort(rall(jest));
		sort(rall(ma_byc));
		
		ll bilans = 0;
		bool ok = true;

		int id_jest = -1;
		int id_ma_byc = -1;

		ll kon_jest = -1;
		ll kon_ma_byc = -1;
		
		ll poz = 0;
		
		while(poz < suma){
			if(poz > kon_jest){
				id_jest++;
				kon_jest += jest[id_jest].s;
			}
			if(poz > kon_ma_byc){
				id_ma_byc++;
				kon_ma_byc += ma_byc[id_ma_byc].s; 
			}
			bilans += (ll)(jest[id_jest].f - ma_byc[id_ma_byc].f) * (min(kon_ma_byc, kon_jest)-poz+1);
			poz = min(kon_ma_byc, kon_jest)+1;
			if(bilans < 0)
				ok = false;
		}
		if(bilans != 0)
			ok = false;
		cout << (ok ? "TAK\n" : "NIE\n");
	}
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);

	solve();
}