// Original C++ solution (slower):
#include <queue>

#include <vector>

#include <iostream>

#pragma warning(disable : 4996)

using namespace std;

void reader(int *x) {

	int k = getchar_unlocked(); *x = k - '0';

	while (1) {

		k = getchar_unlocked();

		if (k < '0' || k > '9') break;

		*x = ((*x) << 1) + ((*x) << 3) + k - '0';

	}

}

int N, M, Q, a[200009], b[200009], r[200009], f[200009], dist[200009], ret[200009]; vector<int> g[200009], gt[200009]; bool vis[100009];

int main() {

	reader(&N), reader(&M), reader(&Q);

	for (int i = 0; i < M; i++) {

		reader(&a[i]), reader(&b[i]), a[i]--, b[i]--;

		g[a[i]].push_back(b[i]);

		g[b[i]].push_back(a[i]);

	}

	fill(f, f + M, -1);

	for (int i = 0; i < Q; i++) reader(&r[i]), r[i]--, f[r[i]] = i;

	queue<int> que1; que1.push(0);

	fill(dist, dist + N, -1); dist[0] = 0;

	while (!que1.empty()) {

		int u = que1.front(); que1.pop();

		for (int i : g[u]) {

			if (dist[i] == -1) {

				dist[i] = dist[u] + 1;

				que1.push(i);

			}

		}

	}

	for (int i = 0; i < M; i++) {

		if (dist[a[i]] == dist[b[i]]) continue;

		if (dist[a[i]] > dist[b[i]]) swap(a[i], b[i]);

		if (f[i] == -1) gt[a[i]].push_back(b[i]);

	}

	int cnt = N - 1;

	vector<bool> vis(N); vis[0] = true;

	queue<int> que2; que2.push(0);

	while (!que2.empty()) {

		int u = que2.front(); que2.pop();

		for (int i : gt[u]) {

			if (!vis[i]) {

				vis[i] = true; cnt--;

				que2.push(i);

			}

		}

	}

	for (int i = Q - 1; i >= 0; i--) {

		ret[i] = cnt;

		if (dist[a[r[i]]] == dist[b[r[i]]]) continue;

		gt[a[r[i]]].push_back(b[r[i]]);

		if (vis[a[r[i]]] && !vis[b[r[i]]]) {

			vis[b[r[i]]] = true; cnt--;

			queue<int> que; que.push(b[r[i]]);

			while (!que.empty()) {

				int u = que.front(); que.pop();

				for (int j : gt[u]) {

					if (!vis[j]) {

						vis[j] = true, cnt--;

						que.push(j);

					}

				}

			}

		}

	}

	for (int i = 0; i < Q; i++) printf("%d\n", ret[i]);

	return 0;

}

// Optimized solution to the same problem (faster):
#include <set>

#include <queue>

#include <vector>

#include <iostream>

#include <algorithm>

#pragma warning(disable : 4996)

using namespace std;

struct edge {

	int to, id;

};

bool operator<(const edge& e1, const edge& e2) {

	if (e1.to != e2.to) return e1.to < e2.to;

	return e1.id < e2.id;

}

int N, M, Q, a[200009], b[200009], dist[100009], r[200009], ans[200009];

vector<int> g[100009]; set<edge> G[100009];

bool ok[200009], vis[200009];

int main() {

	scanf("%d%d%d", &N, &M, &Q);

	for (int i = 0; i < M; i++) {

		scanf("%d%d", &a[i], &b[i]); a[i]--, b[i]--;

		g[a[i]].push_back(b[i]);

		g[b[i]].push_back(a[i]);

	}

	fill(dist + 1, dist + N, 999999999);

	queue<int> que; que.push(0);

	while (!que.empty()) {

		int u = que.front(); que.pop();

		for (int i : g[u]) {

			if (dist[i] == 999999999) {

				dist[i] = dist[u] + 1;

				que.push(i);

			}

		}

	}

	fill(ok, ok + M, true);

	for (int i = 0; i < Q; i++) scanf("%d", &r[i]), ok[--r[i]] = false; r[Q] = M;

	for (int i = 0; i < M; i++) {

		if (dist[a[i]] > dist[b[i]]) swap(a[i], b[i]);

		if (dist[a[i]] != dist[b[i]] && ok[i]) {

			G[a[i]].insert(edge{ b[i], i });

		}

	}

	int sum = 1;

	fill(ok, ok + N, false); ok[0] = true;

	for (int i = Q - 1; i >= 0; i--) {

		if (ok[a[r[i + 1]]]) {

			queue<int> que2; que2.push(a[r[i + 1]]);

			while (!que2.empty()) {

				int u = que2.front(); que2.pop();

				if (!ok[u]) ok[u] = true, sum++;

				vector<int> l;

				for (edge e : G[u]) {

					if (!vis[e.id]) {

						vis[e.id] = true; l.push_back(e.id);

						que2.push(e.to);

					}

				}

				for (int i : l) {

					G[u].erase(edge{ b[i], i });

				}

			}

		}

		ans[i] = sum;

		if (dist[a[r[i]]] != dist[b[r[i]]]) {

			G[a[r[i]]].insert(edge{ b[r[i]], r[i] });

		}

	}

	for (int i = 0; i < Q; i++) printf("%d\n", N - ans[i]);

	return 0;

}

// Fastest known solution for this problem: