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 Scala (2.13.1) (judge verdict: Time Limit Exceeded)

//import java.util.Scanner

import scala.collection.immutable._
import scala.collection.mutable
import scala.language.postfixOps

/**
  * Scala implementation of a faster java.util.Scanner
  * See: http://codeforces.com/blog/entry/7018
  */

object Scanner {
  private val buf = new Array[Byte](1024);
  private var ptr = 0;
  private var len = 0

  @inline private def isPrintableChar(c: Int): Boolean = 33 <= c && c <= 126

  @inline private def hasNextByte(): Boolean = if (ptr >= len) {
    ptr = 0;
    len = System.in.read(buf);
    len > 0
  } else true

  @inline private def hasNext(): Boolean = {
    while (hasNextByte() && !isPrintableChar(buf(ptr))) ptr += 1
    hasNextByte()
  }

  @inline private def readByte(): Byte = if (hasNextByte()) {
    val res = buf(ptr);
    ptr += 1;
    res
  } else -1

  def next(): String = {
    if (!hasNext()) ???
    val sb = new StringBuilder;
    var b = readByte()
    while (isPrintableChar(b)) {
      sb.append(b.toChar);
      b = readByte()
    }
    sb.toString
  }

  def nextInt(): Int = {
    val n = nextLong()
    if (n < Int.MinValue || Int.MaxValue < n) ???
    n.toInt
  }

  def nextLong(): Long = {
    if (!hasNext()) ???
    var minus = false;
    var b = readByte()
    if (b == '-') {
      minus = true;
      b = readByte()
    }

    @scala.annotation.tailrec
    def go(b: Byte, n: Long = 0): Long = if ('0' <= b && b <= '9') go(readByte(), n * 10 + b - '0') else if (minus) -n else n

    go(b)
  }

  def nextDouble(): Double = next().toDouble
}


class IUnionFind(val size: Int) {

  private case class Node(var parent: Option[Int], var treeSize: Int)

  private val nodes = Array.fill[Node](size)(Node(None, treeSize = 1))

  def union(t1: Int, t2: Int): IUnionFind = {
    if (t1 == t2) return this

    val root1 = root(t1)
    val root2 = root(t2)
    if (root1 == root2) return this

    val node1 = nodes(root1)
    val node2 = nodes(root2)

    if (node1.treeSize < node2.treeSize) {
      node1.parent = Some(t2)
      node2.treeSize += node1.treeSize
    } else {
      node2.parent = Some(t1)
      node1.treeSize += node2.treeSize
    }
    this
  }

  def connected(t1: Int, t2: Int): Boolean = t1 == t2 || root(t1) == root(t2)

  def root(t: Int): Int = nodes(t).parent match {
    case None => t
    case Some(p) =>
      nodes(t).parent = Some(root(p))
      nodes(t).parent.get
  }

  def nodeSize(t: Int): Int = nodes(root(t)).treeSize
}


object Main {
  def solve(): Unit = {

    val N, M = Scanner.nextInt()
    val a = Array.fill(N)(List[Int]())
    Array.fill(M)((Scanner.nextInt() - 1, Scanner.nextInt() - 1))
      .foreach(x => {
        if (x._2 != 0) a(x._1) :+= x._2
        if (x._1 != 0) a(x._2) :+= x._1
      })
    val ans: Array[Option[Int]] = Array.fill(N - 1)(None: Option[Int])
    val que = new mutable.Queue[Int]()
    que.enqueue(0)
    while (que.nonEmpty) {
      val now = que.dequeue()
      a(now).foreach(i => if (ans(i - 1).isEmpty) {
        ans(i - 1) = Some(now + 1)
        que.enqueue(i)
      })
    }

    println("Yes")
    println(ans.map(_.get).mkString("\n"))
  }

  def recursive(N: Int,
                A: IndexedSeq[Array[Long]],
                group: Array[Array[Int]] = Array.fill(3)(Array[Int]()),
                ans: Long = 0,
                now: Int = 0): Long = {
    if (now == N) ans
    else (0 until 3).map(i => recursive(N, A, {
      val nextGroup = group.clone()
      nextGroup(i) :+= now
      nextGroup
    }, ans + group(i).map(j => (now, j)).map { case (i, j) => A(Math.min(i, j))(Math.max(i, j) - Math.min(i, j) - 1) }.sum
      , now + 1)).max
  }

  @scala.annotation.tailrec
  def recursive2(N: Long, now: Long = 1): Long = {
    if (N >= now) recursive2(N - now, now + 1)
    else now - 1
  }

  def flip(A: Array[Boolean], B: Array[(Int, Long)]): Array[(Int, Long)] = {
    B.map { case (index, _) =>
      (index, (if (0 < index && A(index - 1)) B(index - 1)._2 else 0L) + (if (index + 1 < B.length && !A(index + 1)) B(index + 1)._2 else 0L))
    }
  }

  val mod: Long = (1e9 + 7).toLong


  @scala.annotation.tailrec
  def factorial(now: Long, ans: Long = 1): Long = {
    if (now == 0) 0 else if (now == 1) ans else factorial(now - 1, now * ans)
  }

  implicit class implicitInt(val N: Int) {
    def times[B](function: Int => B): IndexedSeq[B] = (0 until N).map(function)
  }

  def calc(input: (Long, Long, Long, Long, Long, Long)): Long = {
    input match {
      case (a, b, c, x, y, z) =>
        var ans: Long = Long.MaxValue
        if (b >= 3 && (a >= 2 || c >= 2)) {
          ans = Math.min(ans, Math.max(a - (b - 1), 0) * x + Math.max(c - Math.min(a - 1, b - 2), 0) * z)
          ans = Math.min(ans, Math.max(a - Math.min(c - 1, b - 2), 0) * x + Math.max(c - (b - 1), 0) * z)
        }
        //        println(ans)
        if (b >= 2 && (a >= b && c >= b + 1)) {
          ans = Math.min(ans, y + Math.max(0, a - (c - 1)) * x)
        }
        //        println(ans)
        if (b >= 2 && (a >= b + 1 && c >= b)) {
          ans = Math.min(ans, y + Math.max(0, c - (a - 1)) * z)
        }
        //        println(ans)
        if (a >= 2 && c >= 2)
          if (a != c)
            ans = Math.min(ans, Math.max(0, b - (Math.min(a, c) - 1)) * y)
          else if (a >= 3 && c >= 3)
            ans = Math.min(ans, Math.max(0, b - (Math.min(a, c) - 2)) * y + Math.min(x, z))
        //        println(ans)
        if (ans == Long.MaxValue) ans = -1
        ans
    }
  }

  def time(s: String): Int =
    s.substring(0, 2).toInt * 60 + s.substring(3, 5).toInt

  /*
  def getPermutation(begin: Long = 0): LazyList[Long] =
    LazyList.cons(begin, getPermutation(begin + 1))

  def getFibonacci(prevprev: Long = 0, prev: Long = 1): LazyList[Long] =
    LazyList.cons(prevprev, getFibonacci(prev, prevprev + prev))

  @scala.annotation.tailrec
  def recursive2(X: Set[Long], Y: LazyList[Long]): Long = if (X.contains(Y.head)) X.size else recursive2(X + Y.head, Y.tail)
*/
  def check(i: Int, X: String): Long = {
    if (X == "") 0 else check(i, X.tail) * i + X.head.toString.toLong
  }

  def shift(n: Long): Long = {
    if (n == 0) 0
    else if (n == 1) 1
    else shift(n - 1) << 1
  }

  def unShift(n: Long): Long = {
    if (n == 0) 0
    else unShift(n >> 1) + 1
  }

  @scala.annotation.tailrec
  def gcd(i: Long, j: Long): Long = {
    if (i < j) gcd(j, i)
    else if (j == 0) i
    else gcd(j, i % j)
  }

  def primeFactors(i: Long): List[Long] = primeFactors_(i, 1).sorted


  def primeFactors_(i: Long, j: Long): List[Long] = {
    if (j * j > i) List.empty else if (i % j == 0) primeFactors_(i, j + 1) ++ List[Long](j, i / j) else primeFactors_(i, j + 1)
  }

  def main(args: Array[String]): Unit = {
    solve()
  }
}

/*
object Util {
  def getPermutation(begin: Long = 0): LazyList[Long] =
    LazyList.cons(begin, getPermutation(begin + 1))

  def getPrimeList: LazyList[Long] =
    getPrimeListRecursive(getPermutation(begin = 2))

  private def getPrimeListRecursive(A: LazyList[Long]): LazyList[Long] =
    LazyList.cons(A.head, getPrimeListRecursive(A.tail.filter(_ % A.head != 0)))

  def fib(a: Long = 0, b: Long = 1, mod: Long = Long.MaxValue): LazyList[Long] = a #:: fib(b % mod, (a + b) % mod, mod)
}

 */

object ArabicRoman {

  type =?>[A, B] = PartialFunction[A, B]

  private val codeTable: List[(Int, String)] = List(
    (1000, "M"), (900, "CM"), (500, "D"), (400, "CD"), (100, "C"),
    (90, "XC"), (50, "L"), (40, "XL"), (10, "X"), (9, "IX"), (5, "V"), (4, "IV"), (1, "I"))

  val arabicToRoman: Int =?> String = {
    case src if src >= 1 && src <= 3999 =>

      @scala.annotation.tailrec
      def convert(left: Int, cont: String = "", code: List[(Int, String)] = codeTable): String = {
        val (unitVal, unitChar) = code.head
        left - unitVal match {
          case n if n == 0 => cont + unitChar
          case n if n > 0 => convert(n, cont + unitChar, code)
          case _ => convert(left, cont, code.tail)
        }
      }

      convert(src)
  }

  val romanToArabic: String =?> Int = {
    case src if Option(src).exists { s => {
      s.nonEmpty && """[^MDCLXVI]""".r.findFirstMatchIn(s.toUpperCase).isEmpty
    }
    } =>

      @scala.annotation.tailrec
      def convert(left: String, cont: Int = 0, code: List[(Int, String)] = codeTable): Int = {
        val (unitVal, unitChar) = code.head
        left.splitAt(unitChar.length) match {
          case ("", _) => cont
          case (`unitChar`, tail) => convert(tail, cont + unitVal, code)
          case _ => convert(left, cont, code.tail)
        }
      }

      convert(src.toUpperCase())
  }
}

object PowMod {
  def naivePowMod(a: Long, k: Long, m: Long): Long = {
    var t: Long = 1
    val aMod: Long = a % m

    for (_ <- 1L to k) {
      t = ((t % m) * aMod) % m
    }
    t.toInt
  }

  def powMod(a: Long, k: Long, m: Long): Long = {
    if (k == 0) 1
    else {
      val t = powMod(a, k / 2, m)
      if ((k % 2) == 0) (t * t) % m else (((t * t) % m) * a) % m
    }
  }
}