Problem:
Score: 400 points

Problem Statement

There is a cave.

The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside.

It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage.

Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1.

If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible.

Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 2 \times 10^5

1 \leq A_i, B_i \leq N\ (1 \leq i \leq M)

A_i \neq B_i\ (1 \leq i \leq M)

One can travel between any two rooms by traversing passages.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_M B_M

Output

If there is no way to place signposts satisfying the objective, print No.

Otherwise, print N lines. The first line should contain Yes, and the i-th line (2 \leq i \leq N) should contain the integer representing the room indicated by the signpost in Room i.

Sample Input 1

4 4
1 2
2 3
3 4
4 2

Sample Output 1

Yes
1
2
2

If we place the signposts as described in the sample output, the following happens:

Starting in Room 2, you will reach Room 1 after traversing one passage: (2) \to 1. This is the minimum number of passages possible.

Starting in Room 3, you will reach Room 1 after traversing two passages: (3) \to 2 \to 1. This is the minimum number of passages possible.

Starting in Room 4, you will reach Room 1 after traversing two passages: (4) \to 2 \to 1. This is the minimum number of passages possible.

Thus, the objective is satisfied.

Sample Input 2

6 9
3 4
6 1
2 4
5 3
4 6
1 5
6 2
4 5
5 6

Sample Output 2

Yes
6
5
5
1
1

If there are multiple solutions, any of them will be accepted.

Solution in Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import static java.lang.Math.*;

public class Main {
    public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		Graph g = new Graph(n, false);
		for(int i = 0; i < m; i++){
		    int from = sc.nextInt()-1;
		    int to = sc.nextInt()-1;
		    g.add(from, to);
		}
		
		// 計算
		String result = "No";
		g.bfs(0);
		boolean flg = true;
		for(int i = 1; i < n; i++){
		    if(g.prev[i] == -1) flg = false;
		}
		if(flg) result = "Yes";
		//System.out.println(Arrays.toString(g.prev));
		
		// 出力
		System.out.println(result);
		if(!flg) return;
		for(int i = 1; i < n; i++){
		    System.out.println(g.prev[i]+1);
		}
	}
	
	public static class Scanner {
		private BufferedReader br; private StringTokenizer tok;
		public Scanner(InputStream is) throws IOException { br = new BufferedReader(new InputStreamReader(is));}
		private void getLine() throws IOException { while(!hasNext()) tok = new StringTokenizer(br.readLine());}
		private boolean hasNext() { return tok != null && tok.hasMoreTokens();}
		public String next() throws IOException { getLine(); return tok.nextToken();}
		public int nextInt() throws IOException { return Integer.parseInt(next());}
		public long nextLong() throws IOException { return Long.parseLong(next());}
		public double nextDouble() throws IOException { return Double.parseDouble(next());}
	}
}

class Graph {
	int n;
	boolean directed;
	ArrayList<Edge>[] adjList;
	int m;
	long[] dist;
	long[][] distWF;
	int start;
	int[] prev;
	ArrayList<Integer> path;
	long INF = 1001001001001001001L;

	public Graph (int n, boolean directed) {
		this.n = n;
		this.m = 0;
		this.directed = directed;
		this.adjList = new ArrayList[n];
		for(int i = 0; i < n; i++){
			adjList[i] = new ArrayList<Edge>();
		}
	}
	
	public void add(int from, int to) {
		add(from, to, 1);
	}
	public void add(int from, int to, int w) {
		adjList[from].add(new Edge(from, to, w));
		this.m++;
		if(!directed){
			adjList[to].add(new Edge(to, from, w));
			this.m++;
		}
	}
	
	public long[] bfs(int start) { // O(M)
		this.start = start;
		this.prev = new int[n];
		Arrays.fill(prev, -1);
		prev[start] = start;
		this.dist = new long[n];
		Arrays.fill(dist, INF);
		dist[start] = 0;
		ArrayDeque<Integer> q = new ArrayDeque<Integer>();
		q.add(start);
		while(!q.isEmpty()){
			int now = q.poll();
			for(Edge e : adjList[now]){
				if(dist[e.to] != INF) continue;
				dist[e.to] = dist[now] + e.w;
				prev[e.to] = now;
				q.add(e.to);
			}
		}
		return dist;
	}
	
	public long[] dijkstra(int start) { // O(MlogM)
		this.start = start;
		this.prev = new int[n];
		Arrays.fill(prev, -1);
		prev[start] = start;
		this.dist = new long[n];
		Arrays.fill(dist, INF);
		dist[start] = 0;
		PriorityQueue<Vertex> q = new PriorityQueue<Vertex>((x, y) -> Long.compare(x.distance, y.distance));
		q.add(new Vertex(start, 0));
		while(!q.isEmpty()){
			Vertex now = q.poll();
			if(now.distance != dist[now.id]) continue; 
			for(Edge e : adjList[now.id]){
				long newDist = now.distance + e.w;
				if(dist[e.to] <= newDist) continue;
				dist[e.to] = newDist;
				prev[e.to] = now.id;
				q.add(new Vertex(e.to, newDist));
			}
		}
		return dist;
	}
	
	public long[][] wf() { // O(N^3)
		this.distWF = new long[n][n];
		for(int i = 0; i < n; i++){
			Arrays.fill(distWF[i], INF);
			distWF[i][i] = 0;
		}
		for(ArrayList<Edge> list : adjList){
			for(Edge e : list){
				distWF[e.from][e.to] = e.w;
			}
		}
		for(int k = 0; k < n; k++){
			for(int i = 0; i < n; i++){
				for(int j = 0; j < n; j++){
					distWF[i][j] = min(distWF[i][j], distWF[i][k] + distWF[k][j]);
				}
			}
		}
		return distWF;
	}
	
	public boolean bf(int start) { // O(NM)
		this.start = start;
		this.prev = new int[n];
		Arrays.fill(prev, -1);
		prev[start] = start;
		this.dist = new long[n];
		Arrays.fill(dist, INF);
		dist[start] = 0;
		boolean hasNegativeLoop = false;
		for(int i = 0; i < n; i++){
			for(int j = 0; j < n; j++){
				for(Edge e : adjList[j]){
					if(dist[e.to] <= dist[e.from] + e.w) continue;
					dist[e.to] = dist[e.from] + e.w;
					prev[e.to] = e.from;
					if(i == n-1) hasNegativeLoop = true;
				}
			}
		}
		return hasNegativeLoop;
	}
	
	public ArrayList<Integer> pathTo(int to){
		this.path = new ArrayList<Integer>();
		pathToRec(to);
		return path;
	}
	
	private void pathToRec(int to){
		if(to != start) pathToRec(prev[to]);
		path.add(to);
		return;
	}
	
	public String toString(){
		StringBuilder sb = new StringBuilder();
		for(ArrayList<Edge> list : adjList){
			for(Edge e : list){
				sb.append(e.toString());
				sb.append("\n");
			}
		}
		return sb.toString();
	}
	
	public String distToString(){
		if(dist != null) return Arrays.toString(dist);
		StringBuilder sb = new StringBuilder();
		for(int i = 0; i < n; i++){
			for(int j = 0; j < n; j++){
				if(distWF[i][j] == INF){
					sb.append("from:" + i + " to:" + j + " dist:INF" + "\n");
				}else{
					sb.append("from:" + i + " to:" + j + " dist:" + distWF[i][j] + "\n");
				}
			}
		}
		return sb.toString();
	}
	
	class Edge {
		int from;
		int to;
		int w;
		
		Edge(int from, int to, int w){
			this.from = from;
			this.to = to;
			this.w = w;
		}
		
		public String toString(){
			return "from:" + from + " to:" + to + " w:" + w;
		}
	}
	
	class Vertex {
		int id;
		long distance;
		
		Vertex(int id, long distance){
			this.id = id;
			this.distance = distance;
		}
		
		public String toString(){
			return "id:" + id + " distance:" + distance;
		}
	}
}