vector<int> rustMurderer(int n, vector<vector<int>> roads, int s) {
    /*
     * Write your code here.
     */
    unordered_map<int, unordered_set<int>> graph;
    
    /* Represent an input as a graph based on unordered_... types for
       the maximum insertion/removal performance. */
    for (int i = 1; i <= n; ++i)
        graph[i] = unordered_set<int>{};
    for (const auto& edge : roads) {
        graph[edge[0]].insert(edge[1]);
        graph[edge[1]].insert(edge[0]);
    }
    
    /* A list (set) of the top nodes. A top node is a node that we are currently
       analyse paths from. The top nodes have always the minimum possible distance
       measured as required by the task. */
    unordered_set<int> tops;

    /* The resulting distances. This vector also includes an element for a node
       `s` which will be removed later. */
    vector<int> distances(n);

    int current_distance = 0;
    
    /* Kickstart the process from the given node */
    tops.insert(s);
    distances[s - 1] = current_distance++;

    while (!tops.empty()) {
        unordered_map<int, int> ntops_paths;
    
        /* On each iteration we remove nodes and edges of the graph that
           are already processed, thus each node and edge is effectively
           visited once. */
           
        /* Now find out how many direct (one step) paths are there from the
           current list of top nodes to all nodes of the remaining graph. */

        /* Start with listing all possible nodes. */
        for (const auto& kvp : graph)
            ntops_paths[kvp.first] = 0;

        /* As soon as the direct path found, increment the number of paths
           to the node. */
        for (const auto from : tops) {
            for (const auto to : graph[from]) {
                ntops_paths[to]++;
                graph[to].erase(from);  /* Erase backward edge. */
            }
            ntops_paths.erase(from);  /* Exclude top nodes from further analysis. */
            graph.erase(from);  /* Erase the visited node with all forward edges. */
        }

        /* Now construct a new list of top nodes based on the fact that each node
           that is unreachable from at least one of the current top nodes via a
           "main road" is reachable via "side road". Effectively it means that
           the path number calculated above is at least 1 less than the number of
           the current top nodes. */
        unordered_set<int> ntops;
        for (const auto& kvp : ntops_paths)
            if (kvp.second < tops.size())
                ntops.insert(kvp.first);
        
        /* Populate distances for the new top nodes. */
        for (const auto top : ntops)
            distances[top - 1] = current_distance;
        ++current_distance;
        
        /* Replace the current list of top nodes with the new one. */
        tops = move(ntops);
    }
    
    distances.erase(distances.begin() + s - 1);

    return distances;
}

void solve(){
	ll n,e;
	cin>>n>>e;
	unordered_map<ll,list<ll>> m;
	vector<ll> dis(n+1,INT_MAX);
	queue<ll> q;
	set<ll> l;


	for(ll i=0;i<n;i++){
		l.insert(i+1);
	}

	for(ll i=0;i<e;i++){
		ll a,b;
		cin>>a>>b;
		m[a].push_back(b);
		m[b].push_back(a);
	}
	ll start;
	cin>>start;

	q.push(start);
	l.erase(start);
	dis[start]=0;

	while(!q.empty() and l.size()){
		ll p=q.front();
		q.pop();

		list<ll> ne;

		for(ll x:l){
		if(find(m[p].begin(),m[p].end(),x)==m[p].end()){
				ne.push_back(x);
			}
		}
		for(ll x:ne){
			q.push(x);
			dis[x]=dis[p]+1;
			l.erase(x);
		}
	}
	for(auto a:dis){
		if(a==0 or a==INT_MAX)
			continue;
		cout<<a<<" ";
	}
    cout<<endl;
}