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: Time Limit Exceeded)

import java.io.InputStream;
import java.io.IOException;
import java.io.PrintStream;
import java.math.*;
import java.util.*;

public class Main {

  public static void main(String[] args) {
    solve(System.in, System.out);
  }

  private static class Path {

    public Path(int node, int prev, int hop) {
      this.node = node;
      this.prev = prev;
      this.hop = hop;
    }

    int node;
    int prev;
    int hop;
  }

  static void solve(InputStream is, PrintStream os) {
    Scanner sc = new Scanner(is);

    /* read */
    int n = sc.nextInt();
    int m = sc.nextInt();

    Map<Integer, Set<Integer>> adj = new HashMap<>();
    for (int i = 0; i < n; i++) {
      adj.put(i, new HashSet<>());
    }

    for (int i = 0; i < m; i++) {
      int a = sc.nextInt() - 1;
      int b = sc.nextInt() - 1;

      adj.get(a).add(b);
      adj.get(b).add(a);
    }

    Queue<Path> queue = new ArrayDeque<>();
    queue.add(new Path(0, 0, 0));
    Set<Integer> visited = new HashSet<>();

    int[] prev = new int[n];

    for (int i = 0; i < n; i++) {
      prev[i] = -1;
    }

    while (!queue.isEmpty()) {
      Path last = queue.remove();
      if (prev[last.node] < 0) prev[last.node] = last.prev;

      visited.add(last.node);
      for (int next : adj.get(last.node)) {
        if (visited.contains(next)) continue;
        queue.add(new Path(next, last.node, last.hop + 1));
      }
    }

    if (visited.size() == n) {
      os.println("Yes");

      for (int i = 1; i < n; i++) {
        os.println(prev[i] + 1);
      }
    } else {
      os.println("No");
    }
  }

  private static long gcd(long m, long n) {
    if (m < n) return gcd(n, m);
    if (n == 0) return m;
    return gcd(n, m % n);
  }

  private static long lcm(long m, long n) {
    return m / gcd(m, n) * n;
  }

  private static class Modular {

    private final long mod;

    private Modular(long mod) {
      this.mod = mod;
    }

    private long inverse(long n) {
      return pow(n, mod - 2) % mod;
    }

    private long pow(long x, long n) {
      /*
      long ans = 1;
      while (n > 0) {
        if ((n & 1) == 1) {
          ans = ans * x % mod;
        }
        x = x * x % mod;
        n >>= 1;
      }
      return ans;
      */
      if (n == 0) {
        return 1;
      }
      return n % 2 == 0 ?
          pow(x * x % mod, n / 2) % mod :
          x % mod * pow(x, n - 1) % mod;
    }
  }

  private static class ModularNFixedCombination {

    private final Modular modular;
    private final long[] comb;
    private final long N;

    public ModularNFixedCombination(long N, int size, long mod) {
      this.modular = new Modular(mod);
      this.comb = new long[size + 1];
      this.N = N;

      comb[0] = 1;

      for (int i = 1; i <= size; i++) {
        comb[i] = ((comb[i - 1] * (N - (i - 1))) % mod * modular.inverse(i)) % mod;
      }
    }

    long combination(int n, int k) {
      if (n != N) {
        throw new IllegalStateException();
      }
      return comb[k];
    }

    long repeatedCombination(int n, int k) {
      return combination(n + k - 1, k);
    }
  }

  private static class ModularCombination {

    private final long fact[];
    private final long mod;
    private final Modular modular;

    public ModularCombination(int size, long mod) {
      this.fact = new long[size + 1];
      this.mod = mod;
      this.modular = new Modular(mod);

      this.fact[0] = 1;
      for (int i = 1; i <= size; i++) {
        fact[i] = (fact[i - 1] * i) % mod;
      }
    }

    private long factorial(int n) {
      return fact[n];
    }

    long combination(int n, int k) {
      return
          (
              (
                  (
                      factorial(n) * modular.inverse(factorial(n - k))
                  ) % mod
              ) * modular.inverse(factorial(k))
          ) % mod;
    }

    long repeatedCombination(int n, int k) {
      return combination(n + k - 1, k);
    }
  }

  private static class Scanner {

    private final InputStream is;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    Scanner(InputStream is) {
      this.is = is;
    }

    private boolean hasNextByte() {
      if (ptr < buflen) {
        return true;
      } else {
        ptr = 0;
        try {
          buflen = is.read(buffer);
        } catch (IOException e) {
          e.printStackTrace();
        }
        if (buflen <= 0) {
          return false;
        }
      }
      return true;
    }

    private int readByte() {
      if (hasNextByte()) {
        return buffer[ptr++];
      } else {
        return -1;
      }
    }

    private static boolean isPrintableChar(int c) {
      return 33 <= c && c <= 126;
    }

    public boolean hasNext() {
      while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
        ptr++;
      }
      return hasNextByte();
    }

    public String next() {
      if (!hasNext()) {
        throw new NoSuchElementException();
      }
      StringBuilder sb = new StringBuilder();
      int b = readByte();
      while (isPrintableChar(b)) {
        sb.appendCodePoint(b);
        b = readByte();
      }
      return sb.toString();
    }

    public long nextLong() {
      if (!hasNext()) {
        throw new NoSuchElementException();
      }
      long n = 0;
      boolean minus = false;
      int b = readByte();
      if (b == '-') {
        minus = true;
        b = readByte();
      }
      if (b < '0' || '9' < b) {
        throw new NumberFormatException();
      }
      while (true) {
        if ('0' <= b && b <= '9') {
          n *= 10;
          n += b - '0';
        } else if (b == -1 || !isPrintableChar(b)) {
          return minus ? -n : n;
        } else {
          throw new NumberFormatException();
        }
        b = readByte();
      }
    }

    public int nextInt() {
      long nl = nextLong();
      if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) {
        throw new NumberFormatException();
      }
      return (int) nl;
    }

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