set<int> ap;

void findAP(int u, int par, vector<vector<int>> &G, vector<int> &low, vector<int> &time, int timer){

	time[u] = low[u] = timer;

	// time[u] stores the time at which u was visited, and low[v] stores the minimum time among nodes that 
	// can be reached from v ( minimum of low[i] of all its adjacent vertex except the parent ). u is an articulation
	// point if : 
	// 1) u is root -> It's children are disconnected ( The dfs calls it has to make to its children > 1 )
	// 2) u is not root -> If time[u]<=low[v]
	
	int dfs = 0;
	for(auto &v: G[u]){

		if(v==par) continue;

		if(time[v]==-1){
			dfs++;
			findBridge(v, u, G, low, time, timer+1, bridges);
			if(par!=-1&&time[u]<=low[v])
				ap.insert(u);
			low[u] = min(low[u], low[v]);
		}else{
			// Here, in addition to bridge, a node can also try to touch a neighbor which is not
			// it's parent.
			low[u] = min(low[u], time[v]);
		}

	}
	
	if(par==-1 && dfs>1) ap.insert(u);

}

vector<vector<int>> AP(int n, vector<vector<int>>& E) {

	vector<vector<int>> G(n);
	for(auto &e: E){
		G[e[0]].push_back(e[1]);
		G[e[1]].push_back(e[0]);
	}

	vector<int> low(n, INT_MAX);
	vector<int> time(n, -1);

	// Here time also acts as the visited array
	for(int u=0;u<n;u++){
		if(time[u]==-1)
			findAP(u, -1, G, low, time, 0);
	}

	return ap;

}