#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = array<int, 2>;
#define all(x) begin(x), end(x)
#define sz(x) (int) (x).size()
template<class T> bool smax(T &a, T b) {
return a < b ? a = b, 1 : 0;
}
template<class T> bool smin(T &a, T b) {
return a > b ? a = b, 1 : 0;
}
void solve() {
int n, m, q;
cin >> n >> m >> q;
vector<vector<int>> adj(n);
for (int i = 0; i < m; i++) {
int u, v;
cin >> u >> v;
adj[--v].push_back(--u);
}
const int B = 300;
// this code is sketch!
vector<vector<pii>> best(n);
for (int i = 0; i < n; i++) {
best[i].push_back({i, 0});
for (int j : adj[i]) {
for (auto [idx, len] : best[j]) {
best[i].push_back({idx, len + 1});
}
}
sort(all(best[i]), [](auto &x, auto &y) -> bool {
return x[1] > y[1];
});
best[i].erase(unique(all(best[i])), end(best[i]));
while (sz(best[i]) > B) {
best[i].pop_back();
}
}
// end
vector<bool> ok(n, true);
while (q--) {
int t, y;
cin >> t >> y;
vector<int> c(y);
for (int i = 0; i < y; i++) {
cin >> c[i], ok[--c[i]] = false;
}
--t;
int res = -1;
if (y >= B) {
vector<int> dp(t + 1, -1);
dp[t] = 0;
for (int i = t; i >= 0; i--) {
if (dp[i] == -1) { continue; }
if (ok[i]) { smax(res, dp[i]); }
for (int j : adj[i]) {
smax(dp[j], dp[i] + 1);
}
}
} else {
for (auto [idx, len] : best[t]) {
if (ok[idx]) {
res = len;
break;
}
}
}
for (int i : c) { ok[i] = true; }
cout << res << "\n";
}
}
int main() {
cin.tie(0) -> sync_with_stdio(0);
int t = 1; // cin >> t;
while (t--) solve();
}
/**
* Because of the graph's structure as a DAG,
* we can reverse the graph so that each query is finding
* the furthest town with a beaver that can attend the party.
* Note that because sum(y_i) is bounded, we can block out the queries.
* If y_i >= sqrt(n), then there can be at most sqrt(n) of these queries,
* so we can just directly compute the answer. Otherwise, we can precompute
* the answer. Otherwise, because we have strictly < y_i "unavailable" nodes,
* we only need to compute the sqrt(n) best paths for each node.
*/
