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#include <bits/stdc++.h>

struct DSU{
	DSU () {}
	DSU (int n){ 
		init(n);
	}
	std::vector<int> p, siz;
	void init(int n){
		p.resize(n);
		siz.resize(n);
		for(int i = 0;i < n;i++){
			p[i] = i;
			siz[i] = 1;
		}
	}
	
	int find_par(int u){
		return p[u] = (p[u] == u ? u : find_par(p[u]));
	}
	
	void unite(int a, int b){
		a = find_par(a);
		b = find_par(b);
		if(a == b) return ;
		if(siz[a] < siz[b]) std::swap(a, b);
		p[b] = a;
		siz[a] += siz[b];
	}
};

int main(){
	using namespace std;
	ios::sync_with_stdio(false), cin.tie(nullptr);
	
	int tt;
	cin >> tt;
	
	while(tt--){
		int n, m, k;
		cin >> n >> m >> k; // { nodes, edges, leaders }
		
		vector<int> who_won(n);
		vector<vector<int>> where_won(k);
		for(int i = 0;i < n;i++){
			cin >> who_won[i], who_won[i]--; // kto wygrał w i-tym mieście
			where_won[who_won[i]].push_back(i);
		}
		
		DSU dsu2(n);
		vector<vector<int>> g(n);
		for(int i = 0;i < m;i++){
			int a, b;
			cin >> a >> b, --a, --b;
			g[a].push_back(b);
			g[b].push_back(a);
			dsu2.unite(a, b);
		}
		
		bool ok = true;
		
		for(int i = 0;i < k;i++){
			for(int j = 1;j < (int)where_won[i].size();++j){
				int a = where_won[i][j];
				int b = where_won[i][j - 1];
				if(dsu2.find_par(a) != dsu2.find_par(b)){
					ok = false;
				}
			}
		}
		
		if(!ok){
			cout << "NIE\n";
			continue;
		}

		// we know that every leader stayed in one component as he should
		DSU dsu(n);
		vector<int> done(n);
		for(int i = 0;i < n;i++){
			for(int j : g[i]) if(who_won[i] == who_won[j]){
				dsu.unite(i, j);
			}
		}
		
		vector<int> proc;
		for(int leader = 0;leader < k;leader++){
			if(where_won[leader].empty()) continue;
			int node = where_won[leader][0];
			if(dsu.siz[dsu.find_par(node)] == (int)where_won[leader].size()){
				// this one is done
				for(int x : where_won[leader]){
					done[x] = 1;
				}
				proc.push_back(leader);
			}
		}
		
		if(proc.empty()){
			cout << "NIE\n";
			continue;
		}
		
		vector<vector<int>> collected(k);
		
		while(!proc.empty()){
			int leader = proc.back(); 
			proc.pop_back();
			
			// hard part
			set<int> gr;
			for(int x : where_won[leader]){
				for(int y : g[x]){
					if(!done[y]){
						gr.insert(who_won[y]);
						collected[who_won[y]].push_back(y);
					}
				}
			}
			
			for(int x : gr){
				if(collected[x].size() > 1){
					for(int j = 1;j < (int)collected[x].size();j++){
						int a = collected[x][j];
						int b = collected[x][j - 1];
						dsu.unite(a, b);
					}
					if(dsu.siz[dsu.find_par(collected[x][0])] == (int)where_won[x].size()){
						// we are done with this one.
						for(int x : where_won[x]){
							done[x]=1;
						}
						proc.push_back(x);
					}
				}
				collected[x].clear();
			}
		}
		
		for(int i = 0;i < n;i++){
			ok &= (done[i] == true);
		}
		
		cout << (ok ? "TAK\n" : "NIE\n");
	}
	
	return 0;
}

// graf nie musi być spójny
// start from completed components
// if they are full (all nodes that needed)
// set it as done and add all the new connections that you can have.
// a bit tricky but doable.
// how to implement it as nicely as possible 
// DSU ?