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 11.0.6) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Objects;
import java.util.Queue;
import java.util.StringTokenizer;

public class Main {
	//java11

	public static void main(String[] args) {

		FastScanner sc = new FastScanner(System.in);

		int N = sc.nextInt();
		int M = sc.nextInt();

		Graph g = new Graph(N);
		for(int i=0; i<M; i++) {
			int A = sc.nextInt();
			int B = sc.nextInt();

			g.addEdge(g.getNode(A), g.getNode(B));
			g.addEdge(g.getNode(B), g.getNode(A));
		}

		boolean[] visited = new boolean[N];
		Arrays.fill(visited, false);
		//BFS
		Queue<Node> queue = new ArrayDeque<>();
		queue.add(g.getNode(1));
		while(!queue.isEmpty()) {
			Node n = queue.poll();
			System.out.println("polled: " + n.value);

			//訪問済み
			visited[n.value-1] = true;

			for(Edge edge : n.edges) {
				Node node = edge.to;

				if(!queue.contains(node) && !visited[node.value-1]) {
					queue.add(node);
					System.out.println("added: " + node.value);
					if(node.root != null) {
						System.out.println("No");
						return;
					}else {
						node.root = n;
					}
				}
			}
		}

		System.out.println("Yes");
		for(int i=2; i<=N; i++) {
			System.out.println(g.getNode(i).root.value);
		}

	}

	//一般的なグラフ
	public static class Graph{
		int n;
		Map<Integer, Node> nodes;	//(ノード番号, ノード)

		//1~Nまでをもつグラフ
		public Graph(int n) {
			this.n = n;
			nodes = new HashMap<>(n);

			for(int i=0; i<n; i++) {
				nodes.put(i+1, new Node());
			}
		}

		public Graph(int n, int[] values) {
			this.n = n;
			nodes = new HashMap<>(n);

			for(int i=0; i<n; i++) {
				nodes.put(values[i], new Node(values[i]));
			}
		}

		public Node getNode(int node_n) {
			//ノード番号を指定してノードを取得
			return nodes.get(node_n);
		}

		public void addEdge(Node from, Node to) {
			//重みが1のグラフの場合(重みが均等な場合にも場合によっては使える)
			from.edges.add(new Edge(to, 1));
		}

		public void addEdge(Node from, Node to, long weight) {
			//エッジの重みが異なる場合
			from.edges.add(new Edge(to, weight));
		}

		public void removeEdge(Node from, Node to) {
			//指定したノード間のエッジを削除
			from.edges.remove(new Edge(to, 1));
		}
	}

	public static int count = 0;
	//ノードを表すクラス
	public static class Node{

		int value;
		List<Edge> edges;
		boolean visited = false;
		Node root;

		public Node(){
			value = count+1;
			edges = new LinkedList<Edge>();
			count++;
		}

		public Node(int value) {
			this.value = value;
			edges = new LinkedList<Edge>();
		}

		@Override
		public boolean equals(Object obj) {
			if(obj instanceof Node) {
				Node n = (Node)obj;
				if(this.value == n.value)
					return true;
				else
					return false;
			}

			return false;
		}

		@Override
		public int hashCode() {
			return Objects.hash(this.value);
		}
	}

	//エッジを表すクラス
	public static class Edge{
		Node to;		//行き先
		long weight;		//重み

		public Edge(Node to, long weight) {
			this.to = to;
			this.weight = weight;
		}

		@Override
		public boolean equals(Object obj) {
			if(obj instanceof Edge) {
				Edge e = (Edge)obj;
				if(this.to.equals(e.to))
					return true;
				else
					return false;
			}

			return false;
		}

		@Override
		public int hashCode() {
			return Objects.hash(this.to);
		}
	}

	static class FastScanner {
        private BufferedReader reader = null;
        private StringTokenizer tokenizer = null;

        public FastScanner(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        public String next() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }

            return tokenizer.nextToken("\n");
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public double nextDouble() {
             return Double.parseDouble(next());
         }

        public int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
    }

}