stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<n;i++)

using namespace std;



void reader(int *x) {

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

	while (true) {

		k = getchar();

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

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

	}

}

vector<int>E[100000];

int u[200000], v[200000], r[200000], ans[200000], d[100000], cnt = 1;

bool b[200000], ok[100000];



void bfs(int u) {

	queue<int>que;

	que.push(u);

	while (!que.empty()) {

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

		for (int v : E[p]) {

			if (d[v] == d[p] + 1 && !ok[v]) {

				ok[v] = true; cnt++; que.push(v);

			}

		}

	}

}

int main() {

	int n, m, q; reader(&n); reader(&m); reader(&q);

	rep(i, m) {

		reader(&u[i]); reader(&v[i]);

		u[i]--; v[i]--;

		E[u[i]].push_back(v[i]); E[v[i]].push_back(u[i]);

	}

	queue<int>que;

	memset(d, -1, sizeof(d));

	d[0] = 0; que.push(0);

	while (!que.empty()) {

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

		for (int i : E[p]) {

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

				d[i] = d[p] + 1; que.push(i);

			}

		}

	}

	rep(i, m) {

		if (d[u[i]] > d[v[i]])swap(u[i], v[i]);

	}

	rep(i, n)E[i].clear();

	rep(i, q) {

		reader(&r[i]); r[i]--;

		b[r[i]] = true;

	}

	ok[0] = true;

	rep(i, m) {

		if (!b[i]) {

			if (d[v[i]] - d[u[i]] == 1) {

				E[u[i]].emplace_back(v[i]);

			}

		}

	}

	bfs(0);

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

		ans[i] = n - cnt;

		if (d[v[r[i]]] - d[u[r[i]]] == 1 && !ok[v[r[i]]]) {

			E[u[r[i]]].emplace_back(v[r[i]]);

			if (ok[u[r[i]]]) {

				cnt++; ok[v[r[i]]] = true;

				bfs(v[r[i]]);

			}

		}

	}

	rep(i, q)printf("%d\n", ans[i]);

}

// Faster C++ solution for the same problem (measured speedup: 1.115649534045039x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define S 100000

#define rep(i,n)for(int i=0;i<n;i++)

using namespace std;vector<int>E[S];int u[2*S],v[2*S],r[2*S],ans[2*S],d[S],cnt=1,b[2*S],ok[S]{1};queue<int>que;void bfs(int u, int t) {que.push(u); while (!que.empty()) {int p = que.front(); que.pop(); for (int v : E[p]) {if (t) {if (d[v] > d[p] && !ok[v]) {ok[v] = 1; cnt++; que.push(v);}}else if (!~d[v]) {d[v] = d[p] + 1; que.push(v);}}}}int main(){int n,m,q;scanf("%d%d%d",&n,&m,&q);rep(i, m){scanf("%d%d",&u[i],&v[i]);u[i]--;v[i]--;E[u[i]].push_back(v[i]);E[v[i]].push_back(u[i]);}memset(d,-1,sizeof(d));d[0]=0;bfs(0,0);rep(i,m){if(d[u[i]]>d[v[i]])swap(u[i],v[i]);}rep(i,n)E[i].clear();rep(i,q){scanf("%d",&r[i]);r[i]--;b[r[i]]=1;}rep(i,m){if(!b[i]){if(d[v[i]]>d[u[i]]){E[u[i]].push_back(v[i]);}}}bfs(0,1);for(int i=q-1;i>=0;i--){ans[i]=n-cnt;if(d[v[r[i]]]>d[u[r[i]]]&&!ok[v[r[i]]]){E[u[r[i]]].push_back(v[r[i]]);if(ok[u[r[i]]]){cnt++;ok[v[r[i]]]=1;bfs(v[r[i]],1);}}}rep(i,q)printf("%d\n",ans[i]);}

// Faster C++ solution for the same problem (measured speedup: 1.116645821539032x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<n;i++)

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';

	}

}

vector<int>E[100000];

int u[200000], v[200000], r[200000], ans[200000], d[100000], cnt = 1;

bool b[200000], ok[100000];

queue<int>que;



void bfs(int u, int t) {

	que.push(u);

	while (!que.empty()) {

		int p = que.front(); que.pop(); for (int v : E[p]) {

			if (t) {

				if (d[v] > d[p] && !ok[v]) {

					ok[v] = 1; cnt++; que.push(v);

				}

			}

			else if (!~d[v]) {

				d[v] = d[p] + 1; que.push(v);

			}

		}

	}

}

int main() {

	int n, m, q; reader(&n); reader(&m); reader(&q);

	rep(i, m) {

		reader(&u[i]); reader(&v[i]);

		u[i]--; v[i]--;

		E[u[i]].push_back(v[i]); E[v[i]].push_back(u[i]);

	}

	queue<int>que;

	memset(d, -1, sizeof(d));

	d[0] = 0; bfs(0, 0);

	rep(i, m) {

		if (d[u[i]] > d[v[i]])swap(u[i], v[i]);

	}

	rep(i, n)E[i].clear();

	rep(i, q) {

		reader(&r[i]); r[i]--;

		b[r[i]] = true;

	}

	ok[0] = true;

	rep(i, m) {

		if (!b[i]) {

			if (d[v[i]] - d[u[i]] == 1) {

				E[u[i]].emplace_back(v[i]);

			}

		}

	}

	bfs(0, 1);

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

		ans[i] = n - cnt;

		if (d[v[r[i]]] - d[u[r[i]]] == 1 && !ok[v[r[i]]]) {

			E[u[r[i]]].emplace_back(v[r[i]]);

			if (ok[u[r[i]]]) {

				cnt++; ok[v[r[i]]] = true;

				bfs(v[r[i]], 1);

			}

		}

	}

	rep(i, q)printf("%d\n", ans[i]);

}

// Faster C++ solution for the same problem (measured speedup: 1.1150779404406437x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<n;i++)

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';

	}

}

vector<int>E[100000];

int u[200000], v[200000], r[200000], ans[200000], d[100000], cnt = 1;

bool b[200000], ok[100000];

queue<int>que;



void bfs(int u, int t) {

	que.push(u);

	while (!que.empty()) {

		int p = que.front(); que.pop(); for (int v : E[p]) {

			if (t) {

				if (d[v] > d[p] && !ok[v]) {

					ok[v] = 1; cnt++; que.push(v);

				}

			}

			else if (!~d[v]) {

				d[v] = d[p] + 1; que.push(v);

			}

		}

	}

}

int main() {

	int n, m, q; reader(&n); reader(&m); reader(&q);

	rep(i, m) {

		reader(&u[i]); reader(&v[i]);

		u[i]--; v[i]--;

		E[u[i]].emplace_back(v[i]); E[v[i]].emplace_back(u[i]);

	}

	memset(d, -1, sizeof(d));

	d[0] = 0; bfs(0, 0);

	rep(i, m) {

		if (d[u[i]] > d[v[i]])swap(u[i], v[i]);

	}

	rep(i, n)E[i].clear();

	rep(i, q) {

		reader(&r[i]); r[i]--;

		b[r[i]] = true;

	}

	ok[0] = true;

	rep(i, m) {

		if (!b[i]) {

			if (d[v[i]] - d[u[i]] == 1) {

				E[u[i]].emplace_back(v[i]);

			}

		}

	}

	bfs(0, 1);

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

		ans[i] = n - cnt;

		if (d[v[r[i]]] - d[u[r[i]]] == 1 && !ok[v[r[i]]]) {

			E[u[r[i]]].emplace_back(v[r[i]]);

			if (ok[u[r[i]]]) {

				cnt++; ok[v[r[i]]] = true;

				bfs(v[r[i]], 1);

			}

		}

	}

	rep(i, q)printf("%d\n", ans[i]);

}

// Faster C++ solution for the same problem (measured speedup: 1.1150371349643111x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<n;i++)

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';

	}

}

vector<int>E[100000];

int u[200000], v[200000], r[200000], ans[200000], d[100000], cnt = 1;

bool b[200000], ok[100000]{ 1 };

queue<int>que;



void bfs(int u, int t) {

	que.push(u);

	while (!que.empty()) {

		int p = que.front(); que.pop(); for (int v : E[p]) {

			if (t) {

				if (d[v] > d[p] && !ok[v]) {

					ok[v] = 1; cnt++; que.push(v);

				}

			}

			else if (!~d[v]) {

				d[v] = d[p] + 1; que.push(v);

			}

		}

	}

}

int main() {

	int n, m, q; reader(&n); reader(&m); reader(&q);

	rep(i, m) {

		reader(&u[i]); reader(&v[i]);

		u[i]--; v[i]--;

		E[u[i]].emplace_back(v[i]); E[v[i]].emplace_back(u[i]);

	}

	memset(d, -1, sizeof(d));

	d[0] = 0; bfs(0, 0);

	rep(i, m) {

		if (d[u[i]] > d[v[i]])swap(u[i], v[i]);

	}

	rep(i, n)E[i].clear();

	rep(i, q) {

		reader(&r[i]); r[i]--;

		b[r[i]] = 1;

	}

	rep(i, m) {

		if (!b[i]) {

			if (d[v[i]] > d[u[i]]) {

				E[u[i]].emplace_back(v[i]);

			}

		}

	}

	bfs(0, 1);

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

		ans[i] = n - cnt;

		if (d[v[r[i]]] > d[u[r[i]]] && !ok[v[r[i]]]) {

			E[u[r[i]]].emplace_back(v[r[i]]);

			if (ok[u[r[i]]]) {

				cnt++; ok[v[r[i]]] = 1;

				bfs(v[r[i]], 1);

			}

		}

	}

	rep(i, q)printf("%d\n", ans[i]);

}

// Faster C++ solution for the same problem (measured speedup: 1.1150521101779554x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<n;i++)

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';

	}

}

vector<int>E[100000];

int u[200000], v[200000], r[200000], ans[200000], d[100000], cnt = 1;

char b[200000], ok[100000]{ 1 };

queue<int>que;



void bfs(int u, int t) {

	que.push(u);

	while (!que.empty()) {

		int p = que.front(); que.pop(); for (int v : E[p]) {

			if (t) {

				if (d[v] > d[p] && !ok[v]) {

					ok[v] = 1; cnt++; que.push(v);

				}

			}

			else if (!~d[v]) {

				d[v] = d[p] + 1; que.push(v);

			}

		}

	}

}

int main() {

	int n, m, q; reader(&n); reader(&m); reader(&q);

	rep(i, m) {

		reader(&u[i]); reader(&v[i]);

		u[i]--; v[i]--;

		E[u[i]].emplace_back(v[i]); E[v[i]].emplace_back(u[i]);

	}

	memset(d, -1, sizeof(d));

	d[0] = 0; bfs(0, 0);

	rep(i, m) {

		if (d[u[i]] > d[v[i]])swap(u[i], v[i]);

	}

	rep(i, n)E[i].clear();

	rep(i, q) {

		reader(&r[i]); r[i]--;

		b[r[i]] = 1;

	}

	rep(i, m) {

		if (!b[i]) {

			if (d[v[i]] > d[u[i]]) {

				E[u[i]].emplace_back(v[i]);

			}

		}

	}

	bfs(0, 1);

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

		ans[i] = n - cnt;

		if (d[v[r[i]]] > d[u[r[i]]] && !ok[v[r[i]]]) {

			E[u[r[i]]].emplace_back(v[r[i]]);

			if (ok[u[r[i]]]) {

				cnt++; ok[v[r[i]]] = 1;

				bfs(v[r[i]], 1);

			}

		}

	}

	rep(i, q)printf("%d\n", ans[i]);

}

// Faster C++ solution for the same problem (measured speedup: 1.1150521101779554x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<n;i++)

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';

	}

}

vector<int>E[100000];

int u[200000], v[200000], r[200000], ans[200000], d[100000], cnt = 1;

char b[200000], ok[100000]{ 1 };

deque<int>que;



void bfs(int u, int t) {

	que.emplace_back(u);

	while (!que.empty()) {

		int p = que.front(); que.pop_front();

		for (int v : E[p]) {

			if (t) {

				if (d[v] > d[p] && !ok[v]) {

					ok[v] = 1; cnt++; que.emplace_back(v);

				}

			}

			else if (!~d[v]) {

				d[v] = d[p] + 1; que.emplace_back(v);

			}

		}

	}

}

int main() {

	int n, m, q; reader(&n); reader(&m); reader(&q);

	rep(i, m) {

		reader(&u[i]); reader(&v[i]);

		u[i]--; v[i]--;

		E[u[i]].emplace_back(v[i]); E[v[i]].emplace_back(u[i]);

	}

	memset(d, -1, sizeof(d));

	d[0] = 0; bfs(0, 0);

	rep(i, m) {

		if (d[u[i]] > d[v[i]])swap(u[i], v[i]);

	}

	rep(i, n)E[i].clear();

	rep(i, q) {

		reader(&r[i]); r[i]--;

		b[r[i]] = 1;

	}

	rep(i, m) {

		if (!b[i]) {

			if (d[v[i]] > d[u[i]]) {

				E[u[i]].emplace_back(v[i]);

			}

		}

	}

	bfs(0, 1);

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

		ans[i] = n - cnt;

		if (d[v[r[i]]] > d[u[r[i]]] && !ok[v[r[i]]]) {

			E[u[r[i]]].emplace_back(v[r[i]]);

			if (ok[u[r[i]]]) {

				cnt++; ok[v[r[i]]] = 1;

				bfs(v[r[i]], 1);

			}

		}

	}

	rep(i, q)printf("%d\n", ans[i]);

}

// Faster C++ solution for the same problem (measured speedup: 1.1150277061260905x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<n;i++)

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';

	}

}

vector<int>E[100000];

int u[200000], v[200000], r[200000], ans[200000], d[100000], cnt = 1;

char b[200000], ok[100000]{ 1 };

list<int>que;



void bfs(int u, int t) {

	que.emplace_back(u);

	while (!que.empty()) {

		int p = que.front(); que.pop_front();

		for (int v : E[p]) {

			if (t) {

				if (d[v] > d[p] && !ok[v]) {

					ok[v] = 1; cnt++; que.emplace_back(v);

				}

			}

			else if (!~d[v]) {

				d[v] = d[p] + 1; que.emplace_back(v);

			}

		}

	}

}

int main() {

	int n, m, q; reader(&n); reader(&m); reader(&q);

	rep(i, m) {

		reader(&u[i]); reader(&v[i]);

		u[i]--; v[i]--;

		E[u[i]].emplace_back(v[i]); E[v[i]].emplace_back(u[i]);

	}

	memset(d, -1, sizeof(d));

	d[0] = 0; bfs(0, 0);

	rep(i, m) {

		if (d[u[i]] > d[v[i]])swap(u[i], v[i]);

	}

	rep(i, n)E[i].clear();

	rep(i, q) {

		reader(&r[i]); r[i]--;

		b[r[i]] = 1;

	}

	rep(i, m) {

		if (!b[i]) {

			if (d[v[i]] > d[u[i]]) {

				E[u[i]].emplace_back(v[i]);

			}

		}

	}

	bfs(0, 1);

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

		ans[i] = n - cnt;

		if (d[v[r[i]]] > d[u[r[i]]] && !ok[v[r[i]]]) {

			E[u[r[i]]].emplace_back(v[r[i]]);

			if (ok[u[r[i]]]) {

				cnt++; ok[v[r[i]]] = 1;

				bfs(v[r[i]], 1);

			}

		}

	}

	rep(i, q)printf("%d\n", ans[i]);

}

// Faster C++ solution for the same problem (measured speedup: 1.113339151745266x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

#define mygc(c) (c)=getchar_unlocked()

#define mypc(c) putchar_unlocked(c)

void reader(int& x) { int k, m = 0; x = 0; for (;;) { mygc(k); if (k == '-') { m = 1; break; }if ('0' <= k&&k <= '9') { x = k - '0'; break; } }for (;;) { mygc(k); if (k<'0' || k>'9')break; x = x * 10 + k - '0'; }if (m) x = -x; }

void writer(int x, char c) { int s = 0, m = 0; char f[10]; if (x<0)m = 1, x = -x; while (x)f[s++] = x % 10, x /= 10; if (!s)f[s++] = 0; if (m)mypc('-'); while (s--)mypc(f[s] + '0'); mypc(c); }

template<class T> void writerLn(T x) { writer(x, '\n'); }

#define rep(i,n)for(int i=0;i<n;i++)

vector<int>E[100000];

int u[200000], v[200000], r[200000], ans[200000], d[100000], cnt = 1;

char b[200000], ok[100000]{ 1 };

deque<int>que;



void bfs(int u, int t) {

	que.emplace_back(u);

	while (!que.empty()) {

		int p = que.front(); que.pop_front();

		for (int v : E[p]) {

			if (t) {

				if (d[v] > d[p] && !ok[v]) {

					ok[v] = 1; cnt++; que.emplace_back(v);

				}

			}

			else if (!~d[v]) {

				d[v] = d[p] + 1; que.emplace_back(v);

			}

		}

	}

}

int main() {

	int n, m, q; reader(n); reader(m); reader(q);

	rep(i, m) {

		reader(u[i]); reader(v[i]);

		u[i]--; v[i]--;

		E[u[i]].emplace_back(v[i]); E[v[i]].emplace_back(u[i]);

	}

	memset(d, -1, sizeof(d));

	d[0] = 0; bfs(0, 0);

	rep(i, m) {

		if (d[u[i]] > d[v[i]])swap(u[i], v[i]);

	}

	rep(i, n)E[i].clear();

	rep(i, q) {

		reader(r[i]); r[i]--;

		b[r[i]] = 1;

	}

	rep(i, m) {

		if (!b[i]) {

			if (d[v[i]] > d[u[i]]) {

				E[u[i]].emplace_back(v[i]);

			}

		}

	}

	bfs(0, 1);

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

		ans[i] = n - cnt;

		if (d[v[r[i]]] > d[u[r[i]]] && !ok[v[r[i]]]) {

			E[u[r[i]]].emplace_back(v[r[i]]);

			if (ok[u[r[i]]]) {

				cnt++; ok[v[r[i]]] = 1;

				bfs(v[r[i]], 1);

			}

		}

	}

	rep(i, q)writerLn(ans[i]);

}

// Faster C++ solution for the same problem (measured speedup: 1.1137558430074666x):
#include<bits/stdc++.h>

#define rep(i,n)for(int i=0;i<(n);i++)

using namespace std;

typedef pair<int, int>P;



vector<P>E[100000];

int d[100000], b[100000], cnt[200001], u[200000], v[200000], r[200000], que[100000];

int main() {

	int n, m, q; scanf("%d%d%d", &n, &m, &q);

	rep(i, m) {

		scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; r[i] = q;

	}

	rep(i, q) {

		int a; scanf("%d", &a); r[a - 1] = i;

	}

	rep(i, m) {

		E[u[i]].push_back(P(v[i], r[i])); E[v[i]].push_back(P(u[i], r[i]));

	}

	d[0] = 1; b[0] = q;

	int s = 0, g = 1;

	while (s != g) {

		int p = que[s++];

		rep(i, E[p].size()) {

			P v = E[p][i];

			if (d[v.first] == 0) {

				d[v.first] = d[p] + 1; que[g++] = v.first;

			}

			if (d[v.first] == d[p] + 1)

				b[v.first] = max(b[v.first], min(b[p], v.second));

		}

	}

	rep(i, n)cnt[b[i]]++;

	int ans = 0;

	rep(i, q) {

		ans += cnt[i];

		printf("%d\n", ans);

	}

}