RT, #1#5#7#14#17 WA

#include <cstdio>
#include <queue>
#include <vector>
using namespace std;
#define maxn 100005 
typedef long long ll;
const ll mod = 998244353;
ll a[maxn];
vector < int > G1[maxn];
int degree1[maxn];
void add_edge1(int u, int v) {
	G1[u].push_back(v);
	degree1[v]++;
}
vector < int > G2[maxn];
int degree2[maxn];
void add_edge2(int u, int v) {
	G2[u].push_back(v);
	degree2[v]++;
}
int T[maxn], P[maxn],                     C[maxn], g[maxn][55];
ll                    mul[maxn], V[maxn]                      ; // mul[i] 表示第 i 个函数执行后整个序列被乘了多少次  
int n, m;
void topo1() {
	queue < int > q;
	for (int i = 0; i <= m; i++) {
		if (degree2[i] == 0) {
			q.push(i);
		}
	}
	while (!q.empty()) {
		int u = q.front();
		q.pop();
		for (int v, i = 0; i < (int)G2[u].size(); i++) {
			v = G2[u][i];
			degree2[v]--;
			mul[v] = mul[v] * mul[u] % mod;
			if (degree2[v] == 0) {
				q.push(v);
			}
		}
	}
}
ll cnt_of_calls[maxn]; // 砍掉操作 2 之后, 每个函数等效于调用了多少次 
void topo2() {
	cnt_of_calls[0] = 1;
	queue < int > q;
	for (int i = 0; i <= m; i++) {
		if (degree1[i] == 0) {
			q.push(i);
		}
	}
	while (!q.empty()) {
		int u = q.front();
		q.pop();
		long long now_mul = 1;
		for (int v, i = (int)G1[u].size() - 1; i >= 0; i--) { // 注意这里要倒着遍历边, 因为乘法标记的累计是反向的 
			v = G1[u][i];
			cnt_of_calls[v] = (cnt_of_calls[v] + cnt_of_calls[u] * now_mul) % mod;
			now_mul = now_mul * mul[v] % mod;
			degree1[v]--;
			if (degree1[v] == 0) {
				q.push(v);
			}
		}
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", a + i);
	}
	scanf("%d", &m);
	for (int i = 1; i <= m; i++) {
		scanf("%d", T + i);
		if (T[i] == 1) {
			mul[i] = 1;
			scanf("%d %lld", P + i, V + i);
		} else if (T[i] == 2) {
			scanf("%lld", mul + i);
		} else {
			mul[i] = 1;
			scanf("%d", C + i);
			for (int j = 1; j <= C[i]; j++) {
				scanf("%d", g[i] + j);
				add_edge1(i, g[i][j]);
				add_edge2(g[i][j], i);
			}
		}
	}
	int Q;
	scanf("%d", &Q);
	C[0] = Q;
	for (int f, i = 1; i <= Q; i++) {
		scanf("%d", &f);
		g[0][i] = f;
		mul[0] = 1;
		add_edge1(0, g[0][i]);
		add_edge2(g[0][i], 0);
	}
	topo1();
	topo2();
	for (int i = 1; i <= n; i++) {
		a[i] = a[i] * mul[0] % mod;
	}
	for (int i = 0; i < m; i++) {
		if (T[i] == 1) {
			a[P[i]] = (a[P[i]] + V[i] * cnt_of_calls[i]) % mod;
		}
	}
	for (int i = 1; i <= n; i++) {
		printf("%lld ", a[i]);
	}
	return 0;
}