1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
#include <bits/stdc++.h>
using namespace std;
#define fwd(i, a, n) for (int i = (a); i < (n); i++)
#define rep(i, n) fwd(i, 0, n)
#define all(X) X.begin(), X.end()
#define sz(X) int(size(X))
#define pb push_back
#define eb emplace_back
#define st first
#define nd second
using pii = pair<int, int>; using vi = vector<int>;
using ll = long long; using ld = long double;
#ifdef LOC
auto SS = signal(6, [](int) { *(int *)0 = 0; });
#define DTP(x, y) auto operator << (auto &o, auto a) -> decltype(y, o) { o << "("; x; return o << ")"; }
DTP(o << a.st << ", " << a.nd, a.nd);
DTP(for (auto i : a) o << i << ", ", all(a));
void dump(auto... x) { (( cerr << x << ", " ), ...) << '\n'; }
#define deb(x...) cerr << setw(4) << __LINE__ << ":[" #x "]: ", dump(x)
#else
#define deb(...) 0
#endif

/// We can keep only one find & union by creating white vertices on each edge
/// then on fill we don't need to switch anything to white, just merge all neighbors.
/// It means that our graph is also bipartite.
/// So we never join white and colorful nodes.
/// And colorful nodes also have empty reps so it's free.
void solve() {
    int n,m,k;
    cin >> n >> m >> k;
    int N = n+m;

    vector<int> C(N+1, k+1);
    vector<vi> V(N+1);
    vector<int> counts(k+2);
    vector<vi> where(k+2);
    for(int i = 1; i <= n; i++) {
        cin >> C[i];
        counts[C[i]]++;
        where[C[i]].pb(i);
    }

    for(int i = n + 1; i <= N; i++) {
        int a,b;
        cin >> a >> b;
        V[a].pb(i);
        V[i].pb(a);

        V[b].pb(i);
        V[i].pb(b);
    }

    // Variables for f&u
    vi R(N+1);
    /// representantive neighbor of each color;
    vector<unordered_map<int, int>> reps(N+1);
    vi work;

    

    auto Find = [&R](const auto& self, int x) -> int {
        if (R[x] != x) R[x] = self(self, R[x]);
        return R[x];
    };



    auto Union = [&Find, &R, &reps, &work, &C, &counts, &where](const auto& self, int a, int b) -> void {
        a = Find(Find, a);
        b = Find(Find, b);
        if (a == b) return;
        
        counts[C[a]]--;
        if(counts[C[a]] == 1) { // Won't happen for whites as they start with counts == 0
            work.insert(work.end(), all(where[C[a]]));    
        }
        
        if (reps[a].size() < reps[b].size()) swap(a,b);
        R[b] = a;

        // Won't happen for colorful as they do not have reps
        for(auto [c, v]: reps[b]) {
            if (reps[a].find(c) != reps[a].end()) {
                self(self, reps[a][c], v);
            } else {
                reps[a][c] = v;
            }
        }
        reps[b].clear();
        
    };

    for(int i = 1; i <= N; i++) R[i] = i;
    for(int i = 1; i <= k; i++) {
        if (counts[i] == 1) {
            work.pb(where[i][0]);
        }
    }
    for (int i = n + 1; i <= N; i++) {
        int a = V[i][0];
        int b = V[i][1];
        if (C[a] == C[b]) {
            reps[i][C[a]] = a;
            Union(Union, a, b);
        } else {
            reps[i][C[a]] = a;
            reps[i][C[b]] = b;
        }
    }

    int res = n;
    while(!work.empty()) {
        res--;
        int v = work.back();
        work.pop_back();

        for(int i = 1; i < V[v].size(); i++) {
            Union(Union, V[v][0], V[v][i]);
        }
    }

    cout << ((res == 0) ? "TAK" : "NIE") << "\n";
}

int32_t main() {
	cin.tie(0)->sync_with_stdio(0);
	cout << fixed << setprecision(10);

    int t;
    cin >> t;
    while(t--) solve();
}