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: Wrong Answer)

import java.util.*;
import java.io.*;

public class Main {	
	static Map<Integer,List<Integer>> graph;
	static int[] distance;
	static int[] result;
	private static void solve(){
		//Implement solution here.
		graph = new HashMap<Integer, List<Integer>>();
		
		int N = ni();
		int M = ni();
		
		distance = new int[N + 1];
		result = new int[N + 1];
		
		for(int i = 0; i < M; i++) {
			int A = ni();
			int B = ni();
			if(!graph.containsKey(A)) graph.put(A,new ArrayList<Integer>());
			graph.get(A).add(B);
			if(!graph.containsKey(B)) graph.put(B,new ArrayList<Integer>());
			graph.get(B).add(A);
		}
		
		Queue<Integer> q = new ArrayDeque<Integer>();
		Queue<Integer> qd = new ArrayDeque<Integer>();
		q.add(1);
		qd.add(1);
		while(!q.isEmpty()) {
			int idx = q.poll();
			int dist = qd.poll();
			for(int next: graph.get(idx)) {
				if(distance[next] == 0) {
					distance[next] = dist;
					result[next] = idx;
					q.add(next);
					qd.add(dist + 1);
				} else {
					if(distance[next] < dist) {
						continue;
					} else {
						distance[next] = dist;
						result[next] = idx;
						q.add(next);
						qd.add(dist + 1);
					}
				}
			}
		}

		boolean ok = true;
		for(int i = 2; i <= N; i++) {
			if(result[i] == 0) {
				ok = false;
			}
		}
		
		if(ok) {
			for(int i = 2; i <= N; i++) {
				System.out.println("Yes");
				System.out.println(result[i]);
			}
		} else {
			System.out.println("No");
		}
	}
	
//	private static void rec(int from, int idx,int dist) {
//		if(distance[idx] == 0) {
//			distance[idx] = dist;
//			result[idx] = from;
//		} else {
//			if(distance[idx] < dist) {
//				return;
//			} else {
//				distance[idx] = dist;
//				result[idx] = from;
//			}
//		}
//		for(int next:graph.get(idx)) {
//			rec(idx, next,dist + 1);
//		}		
//	}

	//Switch input source (stdin/file)
	private static BufferedReader br;
	public static void main(String[] args){
		br = new BufferedReader(new InputStreamReader(System.in));
		String debugDataPath = System.getenv("DD_PATH");        
		if(debugDataPath != null){
			try{
				br = new BufferedReader(new InputStreamReader(new FileInputStream(new File(debugDataPath))));
			}catch(Exception e){
				throw new RuntimeException(e);
			}
		}
		boolean hasNext = true;
		while(hasNext){
			solve();	
			try{
				br.mark(1);
				if(br.read() > 0){
					br.reset();			
					System.out.println("---");
				} else {
					hasNext = false;
				}
			}catch(Exception e){
				throw new RuntimeException(e);
			} 
		}
	}

	//Input read utility
	private static StringTokenizer tokenizer = null;
	public static String next() {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}
	//Get next single int
	private static int ni() {
		return Integer.parseInt(next());
	}
	//Get next single long
	private static long nl() {
		return Long.parseLong(next());
	}
	//Get next single double
	private static double nd() {
		return Double.parseDouble(next());
	}
	//Get next int array from one line
	private static int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}
	//Get next char array from one line
	private static char[] ns() {
		return next().toCharArray();
	}
	//Get next long array from one line
	private static long[] nal(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}
}