// #pragma GCC target("avx2") // #pragma GCC optimize("O3") // #pragma GCC optimize("unroll-loops") #include #define rep(i, n) for (int i = 0; i < (int)n; ++i) #define rrep(i, n) for (int i = (int)n - 1; i >= 0; --i) #define ALL(v) v.begin(), v.end() #define pb push_back #define eb emplace_back #define fi first #define se second using namespace std; using ll = long long; using ld = long double; const array YESNO = {"NO", "YES"}; const array YesNo = {"No", "Yes"}; const array yesno = {"no", "yes"}; void YES(bool b = true) { cout << YESNO[b] << '\n'; } void Yes(bool b = true) { cout << YesNo[b] << '\n'; } void yes(bool b = true) { cout << yesno[b] << '\n'; } template inline bool chmax(T& a, const U& b) { if (a < b){ a = b; return true; } return false; } template inline bool chmin(T& a, const U& b) { if (a > b) { a = b; return true; } return false; } template void UNIQUE(vector& v) { sort(v.begin(), v.end()); v.erase(unique(v.begin(), v.end())); } template int lb(const vector v, T x) { return distance(v.begin(), lower_bound(v.begin(), v.end(), x)); } template int ub(const vector v, T x) { return distance(v.begin(), upper_bound(v.begin(), v.end(), x)); } /** * @brief 多次元 vector の作成 * @author えびちゃん */ namespace detail { template auto make_vec(vector& sizes, T const& x) { if constexpr (N == 1) { return vector(sizes[0], x); } else { int size = sizes[N-1]; sizes.pop_back(); return vector(size, make_vec(sizes, x)); } } } template auto make_vec(int const(&sizes)[N], T const& x = T()) { vector s(N); for (int i = 0; i < N; ++i) s[i] = sizes[N-i-1]; return detail::make_vec(s, x); } template ostream& operator<<(ostream& os, const vector& v) { for (auto it = v.begin(); it != v.end(); ++it) { if (it == v.begin()) os << *it; else os << ' ' << *it; } return os; } template ostream& operator<<(ostream& os, const pair& p) { os << p.first << ' ' << p.second; return os; } __attribute__((constructor)) void fast_io() { ios::sync_with_stdio(false); cin.tie(nullptr); } // #include // using namespace atcoder; // using mint = modint1000000007; // using mint = modint998244353; void solve() { ll k, z, c; cin >> k >> z >> c; vector a(k), p(k); rep(i, k) cin >> a[i]; rep(i, k) cin >> p[i]; ll g = 0; rep(i, k) g ^= (a[i] % (p[i] + 1)); bool zwin = false; if (g == 0) { if (z > c) { zwin = true; } else { zwin = false; } } else { if (c < g) { zwin = true; } else { // c >= g if (c >= z + g) { zwin = false; } else { zwin = true; } } } cout << (zwin ? 'Z' : 'C') << '\n'; } int main() { int t; cin >> t; while (t--) solve(); }