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 Kotlin (1.3.71) (judge verdict: Accepted)

import java.io.BufferedReader
import java.io.InputStream
import java.io.InputStreamReader
import java.util.*
import kotlin.math.abs
import kotlin.math.max
import kotlin.math.min

val MOD = 1_000_000_007L
/**
 * (dist, parent, queue)
 */
fun traceBfs(g: Array<IntArray>, rt: Int = 0): Array<IntArray> {
  val n = g.size
  val q = IntArray(n)
  val d = IntArray(n)
  val p = IntArray(n){-2}
  var cur = 0
  var last = 1
  p[rt] = -1
  q[0] = rt
  d[rt] = 0
  while (cur < last) {
    val v = q[cur++]
    for (u in g[v]) {
      if (p[u] == -2) {
        p[u] = v
        q[last++] = u
        d[u] = d[v] + 1
      }
    }
  }
  return arrayOf(d, p, q)
}

fun packUGraph(n: Int, from: IntArray, to: IntArray): Array<IntArray> {
  val p = IntArray(n)
  val m = from.size
  for (i in 0 until m) {
    ++p[from[i]]
    ++p[to[i]]
  }
  val g = Array(n){IntArray(p[it])}
  for (i in 0 until m) {
    g[from[i]][--p[from[i]]] = to[i]
    g[to[i]][--p[to[i]]] = from[i]
  }
  return g
}
class Solver(stream: InputStream, private val out: java.io.PrintWriter) {
  fun solve() {
    val N = ni()
    val M = ni()
    val (from, to) = na2(M, -1)
    val g = packUGraph(N, from, to)
    val (_, par, _) = traceBfs(g)
    out.println("Yes")
    for (i in 1 until N) {
      out.println(par[i] + 1)
    }
  }













































  private val isDebug = try {
    // なんか本番でエラーでる
    System.getenv("MY_DEBUG") != null
  } catch (t: Throwable) {
    false
  }

  private var tokenizer: StringTokenizer? = null
  private val reader = BufferedReader(InputStreamReader(stream), 32768)
  private fun next(): String {
    while (tokenizer == null || !tokenizer!!.hasMoreTokens()) {
      tokenizer = StringTokenizer(reader.readLine())
    }
    return tokenizer!!.nextToken()
  }

  private fun ni() = next().toInt()
  private fun nl() = next().toLong()
  private fun ns() = next()
  private fun na(n: Int, offset: Int = 0): IntArray {
    return map(n) { ni() + offset }
  }
  private fun nal(n: Int, offset: Int = 0): LongArray {
    val res = LongArray(n)
    for (i in 0 until n) {
      res[i] = nl() + offset
    }
    return res
  }

  private fun na2(n: Int, offset: Int = 0): Array<IntArray> {
    val a  = Array(2){IntArray(n)}
    for (i in 0 until n) {
      for (e in a) {
        e[i] = ni() + offset
      }
    }
    return a
  }

  private inline fun map(n: Int, f: (Int) -> Int): IntArray {
    val res = IntArray(n)
    for (i in 0 until n) {
      res[i] = f(i)
    }
    return res
  }

  private inline fun debug(msg: () -> String) {
    if (isDebug) System.err.println(msg())
  }

  private fun debug(a: LongArray) {
    debug { a.joinToString(" ") }
  }

  private fun debug(a: IntArray) {
    debug { a.joinToString(" ") }
  }

  private fun debug(a: BooleanArray) {
    debug { a.map { if (it) 1 else 0 }.joinToString("") }
  }

  private fun debugDim(A: Array<IntArray>) {
    if (isDebug) {
      for (a in A) {
        debug(a)
      }
    }
  }

  /**
   * 勝手にimport消されるのを防ぎたい
   */
  private fun hoge() {
    min(1, 2)
    max(1, 2)
    abs(-10)
  }
}

fun main() {
  val out = java.io.PrintWriter(System.out)
  Solver(System.`in`, out).solve()
  out.flush()
}