#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN = 100005;
const int B = 320; // Kvadrat ildiz chegarasi
vector<int> adj[MAXN];
vector<pair<int, int>> best[MAXN]; // {masofa, shahar}
int dist[MAXN], busy[MAXN];
// Ikkita saralangan ro'yxatni birlashtirib, eng yaxshi B tasini saqlash
vector<pair<int, int>> merge(const vector<pair<int, int>>& v1, const vector<pair<int, int>>& v2) {
vector<pair<int, int>> res;
int i = 0, j = 0;
static bool used[MAXN];
while (res.size() < B && (i < v1.size() || j < v2.size())) {
pair<int, int> next;
if (i < v1.size() && (j == v2.size() || v1[i].first > v2[j].first + 1)) {
next = v1[i++];
} else {
next = {v2[j].first + 1, v2[j].second};
j++;
}
if (!used[next.second]) {
used[next.second] = true;
res.push_back(next);
}
}
for (auto& p : res) used[p.second] = false;
return res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int N, M, Q;
cin >> N >> M >> Q;
for (int i = 0; i < M; i++) {
int u, v;
cin >> u >> v;
adj[v].push_back(u); // Teskari yo'nalishda saqlaymiz
}
// 1. Oldindan DP orqali har bir shahar uchun eng yaxshi B ta masofani hisoblash
for (int i = 1; i <= N; i++) {
best[i].push_back({0, i});
for (int prev : adj[i]) {
best[i] = merge(best[i], best[prev]);
}
}
// 2. So'rovlarni qayta ishlash
while (Q--) {
int T, Y;
cin >> T >> Y;
vector<int> C(Y);
for (int i = 0; i < Y; i++) {
cin >> C[i];
busy[C[i]] = 1;
}
if (Y < B) {
// Kam bandlar: Tayyor ro'yxatdan qidiramiz
int ans = -1;
for (auto& p : best[T]) {
if (!busy[p.second]) {
ans = p.first;
break;
}
}
cout << ans << "\n";
} else {
// Ko'p bandlar: DP orqali hisoblaymiz
vector<int> dp(T + 1, -1);
dp[T] = 0;
int ans = -1;
for (int i = T; i >= 1; i--) {
if (dp[i] == -1) continue;
if (!busy[i]) ans = max(ans, dp[i]);
for (int next : adj[i]) {
dp[next] = max(dp[next], dp[i] + 1);
}
}
cout << ans << "\n";
}
// Bandlarni tozalash
for (int x : C) busy[x] = 0;
}
return 0;
}
