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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"io"
	"math"
	"os"
	"strconv"
	"strings"

)


func main() {
	solve(os.Stdin, os.Stdout)
}

type ReaderEx struct {
	baseScanner *bufio.Scanner
	words       []string
	nextWordIdx func() int
	initWordIdx func(int)
}

func NewScannerEx(stdin io.Reader) *ReaderEx {
	var this ReaderEx
	var idx int
	var wordsCnt int
	this.baseScanner = bufio.NewScanner(stdin)
	this.baseScanner.Buffer(make([]byte, 0), 1E+11)
	this.nextWordIdx = func() int {
		cur := idx
		idx = (idx + 1) % wordsCnt
		return cur
	}
	this.initWordIdx = func(wc int) {
		idx = 0
		wordsCnt = wc
	}
	return &this
}

func (me *ReaderEx) Scan(doSplit bool) (isScanned bool) {
	isScanned = me.baseScanner.Scan()
	if doSplit {
		me.words = strings.Fields(me.baseScanner.Text())
	} else {
		me.words = []string{me.baseScanner.Text()}
	}
	me.initWordIdx(len(me.words))
	return
}

func (me *ReaderEx) nextBytes() []byte {
	return []byte(me.words[me.nextWordIdx()])
}
func (me *ReaderEx) nextString() string {
	return me.words[me.nextWordIdx()]
}

type StringArray []string

func (me *ReaderEx) nextStringArray() StringArray {
	return me.words[me.nextWordIdx():]
}

func (me StringArray) toStrings() []string {
	return me
}

func (me StringArray) toInts() []int {
	strValues := me.toStrings()
	values := make([]int, len(strValues))
	for i, v := range strValues {
		n, err := strconv.Atoi(v)
		if err != nil {
			panic(err)
		}
		values[i] = n
	}
	return values
}
func (me *ReaderEx) nextInt() int {
	n, err := strconv.Atoi(me.words[me.nextWordIdx()])

	defer func() {
		if err != nil {
			panic(fmt.Sprintf("getInt(%d)", n))
		}
	}()
	return n
}

func pow(x, y, mod int) (ans int) {

	ans = 1
	for bit := 1 << uint(math.Floor(math.Log2(float64(y)))); ; bit >>= 1 {
		if y&bit != 0 {
			ans *= x
			if mod > 0 {
				ans %= mod
			}
		}
		if bit <= 1 {
			return
		}
		ans *= ans
		if mod > 0 {
			ans %= mod
		}
	}
}

type factType struct {
	prime int
	count int
}

func fact(n int) (facts []*factType) {

	for i := 2; i*i <= n; i++ {
		if n%i == 0 {
			fa := factType{prime: i}
			for n%fa.prime == 0 {
				n /= fa.prime
				fa.count++
			}
			facts = append(facts, &fa)
		}
	}
	if n > 1 {
		facts = append(facts, &factType{prime: n, count: 1})
	}
	return facts
}
func dec2bin(n int) (ret []bool) {
	for bit := 1 << uint(math.Floor(math.Log2(float64(n)))); ; bit >>= 1 {
		ret = append(ret, n&bit != 0)

		if bit <= 1 {
			return
		}
	}

}

func abs(v int) int {
	if v < 0 {
		v = -v
	}
	return v
}

func b2i(b bool) (i int) {
	if b {
		i = 1
	} else {
		i = 0
	}
	return
}

//Edge is
type Edge struct {
	node   *Node
	Bridge *Bridge
}

//Node is
type Node struct {
	id    int
	edges []*Edge
	preid int
}

//Bridge is
type Bridge struct {
	edges [2]Edge
}

//NewBridge is
func NewBridge(a, b *Node) (bridge *Bridge) {
	bridge = &Bridge{}
	bridge.edges = [...]Edge{{node: a, Bridge: bridge}, {node: b, Bridge: bridge}}
	a.edges = append(a.edges, &bridge.edges[1])
	b.edges = append(b.edges, &bridge.edges[0])
	return
}

//Another is
func (edge *Edge) Another() (another *Edge) {
	Bridge := edge.Bridge
	if &Bridge.edges[0] == edge {
		another = &Bridge.edges[1]
	} else {
		another = &Bridge.edges[0]
	}
	return
}

func nodeRead() {
	// for i := 1; i <= n; i++ {
	// 	nodes[i] = Node{id: i}
	// }

	// for i, end := 1, n-1; i <= end; i++ {
	// 	r.Scan(true)
	// 	NewBridge(&nodes[r.nextInt()], &nodes[r.nextInt()])
	// }

}

func solve(stdin io.Reader, stdout io.Writer) {

	r := NewScannerEx(stdin)
	r.Scan(true)
	n, m := r.nextInt(), r.nextInt()

	nodes := make([]*Node, n+1)
	for i := 1; i <= n; i++ {
		nodes[i] = &Node{id: i, preid: -1}
	}
	for i := 0; i < m; i++ {
		r.Scan(true)
		a, b := r.nextInt(), r.nextInt()
		NewBridge(nodes[a], nodes[b])
	}

	type stackInfo struct {
		node *Node
	}
	stack := make([]*stackInfo, 0, n)
	nodes[1].preid = 0
	stack = append(stack, &stackInfo{node: nodes[1]})
	for len(stack) > 0 {
		cur := stack[len(stack)-1]
		stack = stack[:len(stack)-1]

		for _, neither := range cur.node.edges {
			if neither.node.preid < 0 {
				neither.node.preid = cur.node.id
				stack = append(stack, &stackInfo{node: neither.node})
			}
		}
	}
	ans := "Yes"
	for i := 2; i <= n; i++ {
		if nodes[i].preid <= 0 {
			ans = "No"
			break
		}
	}
	fmt.Fprintln(stdout, ans)
	for i := 2; i <= n; i++ {
		fmt.Fprintln(stdout, nodes[i].preid)
	}
}