#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = double;

constexpr int N = 2e3 + 10;
constexpr int Q = 1e6 + 10;
int read() {
    char u, v;
    cin >> u >> v;
    return (u - 48) * 50 + (v - 48);
}
int f[N];
int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }
void merge(int x, int y) {
    int fx = find(x), fy = find(y);
    if (fx == fy) return;
    f[fx] = fy;
}
int n, m, q, ans[Q];
vector<array<int, 3>> qes[N];
vector<pair<int, int>> add_edge[N];
vector<int> edge[N], scc_edge[N];
int step, top;
int dfn[N], low[N], stk[N], instk[N], root[N];
void clear(int opt) {
    step = top = 0;
    memset(dfn, 0, sizeof(int) * (n + 2));
    for (int i = 1; i <= n; i++) scc_edge[i].clear();
    if (opt) return;
    for (int i = 0; i <= m; i++) qes[i].clear();
    for (int i = 1; i <= n; i++) edge[i].clear();
    for (int i = 0; i <= m; i++) add_edge[i].clear();
}
void tarjan(int u) {
    dfn[u] = low[u] = ++step;
    stk[++top] = u;  // 节点 u 入栈
    instk[u] = true; // 标记 u 在栈中
    for (auto v : edge[u]) {
        if (dfn[v] == 0) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (instk[v]) low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
        while (true) {
            int cur = stk[top--]; // 节点 cur 出栈
            instk[cur] = false;   // 节点 cur 出栈
            root[cur] = u;	      // 节点 cur 所在强连通分量根节点 u
            if (cur == u) break;
        }
    }
}
void solve() {
    cin >> n >> m >> q;
    clear(0);
    for (int i = 1; i <= m; i++) {
        vector<int> p(n + 1), cnt(n + 1);
        for (int j = 1; j <= n; j++) {
            p[j] = read();
            cnt[p[j]]++;
        }
        int two = -1, zero = -1, ok = true;
        for (int j = 1; j <= n; j++) {
            if (cnt[j] == 2) {
                if (two == -1) two = j;
                else ok = false;
            } else if (cnt[j] == 0) {
                if (zero == -1) zero = j;
                else ok = false;
            }
        }
        if (!ok) continue; // 该按钮对应的转移边不合法
        if (two == -1) {
            for (int j = 1; j <= n; j++) {
                add_edge[i].push_back({j, p[j]});
            }
        } else for (int j = 1; j <= n; j++) {
            if (cnt[p[j]] == 2) {
                add_edge[i].push_back({j, zero});
            }
        }
    }
    // 询问仅使用前 c 个按钮，能否使得节点 a 可以达到节点 b
    for (int i = 1; i <= q; i++) {
        int a = read(), b = read(), c = read();
        qes[c].push_back({a, b, i});
    }
    // 并查集初始化，用来记录每个强连通分量的根节点
    for (int i = 1; i <= n; i++) f[i] = i;
    for (int c = 0; c <= m; c++) {
        // 反向加边 v -> u
        for (auto [u, v] : add_edge[c]) {
            edge[find(v)].push_back(find(u));
        }
        // 强连通分量缩点 tarjan 算法
        for (int i = 1; i <= n; i++) {
            if (!dfn[i]) tarjan(i);
        }
        vector<int> deg(n + 1);
        set<pair<int, int>> st;
        for (int i = 1; i <= n; i++) {
            for (auto j : edge[i]) {
                int u = root[i], v = root[j];
                if (u != v && !st.count({u, v})) {
                    st.insert({u, v});
                    scc_edge[u].push_back(v);
                    deg[v]++;
                }
            }
        }
        // 将每个节点 i 合并到所在缩点集合 root[i] 中，更新强连通分量缩点建图
        for (int i = 1; i <= n; i++) {
            merge(i, root[i]);
            edge[i] = scc_edge[i];
        }
        // bitset 拓扑排序，求每个点的可达向量
        vector<int> que;
        vector<bitset<N>> dp(n + 1);
        for (int i = 1; i <= n; i++) {
            if (!deg[i]) {
                dp[i][i] = 1;
                que.push_back(i);
            }
        }
        for (int i = 0; i < que.size(); i++) {
            int u = que[i];
            for (auto v : edge[u]) {
                dp[v] |= dp[u];
                if (--deg[v] == 0) {
                    dp[v][v] = 1;
                    que.push_back(v);
                }
            }
        }
        // dp[find(a)][find(b)] 表示原图中 a 是否能够到达 b
        for (auto [a, b, qid] : qes[c]) {
            ans[qid] = dp[find(a)][find(b)];
        }
        clear(1);
    }
    for (int i = 1; i <= q; i++) cout << ans[i];
    cout << "\n";
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int tt = 1;
    cin >> tt;
    while (tt--) {
        solve();
    }
    return 0;
}