#include<bits/stdc++.h>
using namespace std;
#define int long long
using ii = pair<int, int>;
vector<vector<ii>> adj;
int n, m, u;
/* observation: abs(dist(a, x) - dist(b, x)) <= dist(a, b) <= dist(a, x) + dist(b, x)
* in fact, only the left half is useful.
* The answer is YES, iff we can find some abs(dist(a, x) - dist(b, x)) <= p (the query value).
*/
bool dfs(int u, int t, int p, vector<bool>& visited) {
if (visited[u]) return false;
if (u == t) return true;
visited[u] = true;
for (auto [v, w] : adj[u]) {
if (w > p) continue;
bool r = dfs(v, t, p, visited);
if (r) return true;
}
return false;
}
signed main() {
ios_base::sync_with_stdio(false); cin.tie(nullptr);
cin >> n >> m >> u;
// all nodes are 1-indexed.
adj = vector<vector<ii>>(n+1, vector<ii>());
for (int i = 0; i < m; i++) {
int a, b, t; cin >> a >> b >> t;
adj[a].push_back({b, t});
adj[b].push_back({a, t});
}
// queries
while (u--) {
int a, b, p; cin >> a >> b >> p;
vector<bool> visited(n+1, false);
auto ok = dfs(a, b, p, visited);
if (ok) {
cout << "TAIP\n";
}
else {
cout << "NE\n";
}
}
}
