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 Fortran (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

      MODULE intQueueMod
      implicit none
! Module for QUEUE : First in, First out
! reference : プログラミングコンテスト攻略のための
!              アルゴリズムとデータ構造

      integer(16),parameter :: binSize = 10**6
      integer(16),parameter :: precision = 16
      
      type intQueue
        private
        integer(precision) :: bin( 1:binSize ) = 0
        integer(precision) :: head = 1,   tail = 1
        !head : Where the head value is.
        !tail : Where the blank next to tail is.
        contains
        procedure,public,pass :: initialize => intInit
        procedure,public,pass :: isEmpty    => intIsEmpty
        procedure,public,pass :: isFull     => intIsFull
        procedure,public,pass :: enQueue    => intEnQ
        procedure,public,pass :: deQueue    => intDeQ
      end type
      
      contains
!Reset queue. Not nessesary in first time.
      pure subroutine intInit( q )
        implicit none
        class(intQueue),intent(inout) :: q
        
        q%head = 1
        q%tail = 1
      end subroutine intInit
      
!Check status
      pure function intIsEmpty( q ) result( ans )
        implicit none
        class(intQueue),intent(in) :: q
        logical :: ans
        
        ans = q%head==q%tail
      end function intIsEmpty
      
      
      pure function intIsFull( q ) result( ans )
        implicit none
        class(intQueue),intent(in) :: q
        logical :: ans
        
        ans = q%head==mod( q%tail+1,binSize )
      end function intIsFull

!set x to tail
      subroutine intEnQ( q,x )
        implicit none
        class(intQueue)      ,intent(inout) :: q
        integer(precision),intent(in   ) :: x
        
        if( q%isFull() )then
          print*,'Queue is full,Error!';stop
        else
          q%bin(q%tail) = x
          if( q%tail==binSize )then
            q%tail = 1
          else
            q%tail = q%tail + 1
          end if
        end if
      end subroutine intEnQ
      
!get x from head
      impure function intDeQ( q ) result( x )
        implicit none
        class(intQueue),intent(inout) :: q
        integer(precision) :: x
        
        if( q%isEmpty() )then
          print*,'Stack is empty, Error!';stop
        else
          x = q%bin(q%head)
          if( q%head==binSize )then
            q%head = 1
          else
            q%head = q%head + 1
          end if
        end if
      end function intDeQ
      END MODULE intQueueMod
      
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
      MODULE graphMod
      implicit none
! 重みなしのグラフを扱うためのモジュール

      integer(16),parameter,private :: prec = 16
      
      type :: nodes
        integer(prec),allocatable :: next(:)
        integer(prec) :: degree = 0
        contains
        procedure,public,pass :: appendNode
      end type
      
      contains
      subroutine appendNode( node, nodeNumber )
        implicit none
        class(nodes),intent(inout) :: node
        integer(prec),intent(in)  :: nodeNumber
        integer(prec),allocatable :: buffer(:)
        
        if(.not.(allocated(node%next)))then
          allocate(node%next(1))
          node%degree = 0
        end if
        
        if( node%degree==0 )then
          node%next(1) = nodeNumber
          node%degree = 1
          return
        end if
        
        node%degree = node%degree + 1
        buffer = node%next(1:node%degree-1)
        if( size(node%next)<node%degree )then
          deallocate( node%next )
          allocate( node%next(node%degree*2) )
        end if
        node%next(1:node%degree) = [buffer(1:node%degree-1), nodeNumber]
        deallocate( buffer )
      end subroutine appendNode
      
      END MODULE graphMod
      
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!



      module piyoMod
      use intQueueMod
      use graphMod
      
      integer(16),allocatable :: flag(:)
      type(intQueue) :: nextQ,beforeQ
      
      contains
      subroutine solve( rIn,roomNum,before )
        implicit none
        type(nodes),intent(in) :: rIn(:)
        integer(16),intent(in) :: roomNum,before
        integer(16) :: i
        
        if( flag(roomNum)/=-1 ) return
        flag( roomNum ) = before
        if( rIn(roomNum)%degree==0 ) return
        
        do i = 1,rIn(roomNum)%degree
          call nextQ%enQueue( rIn(roomNum)%next(i) )
          call beforeQ%enQueue( roomNum )
        end do
        
      end subroutine
      end module
      
      PROGRAM doubleDots
      use piyoMod
      IMPLICIT NONE
      integer(16) :: n,m,numNei,buffer(5*10**5+1)
      integer(16),allocatable :: a(:),b(:)
      type(nodes),allocatable :: room(:)
      integer(16) :: i
      
      read*,n,m
      allocate( a(m),b(m),room(n),flag(n) )
      flag(1:n) = -1
      do i = 1,m
        read*,a(i),b(i)
      end do
      
      do i = 1,m
        call room(a(i))%appendNode( b(i) )
        call room(b(i))%appendNode( a(i) )
      end do
      
      
      
 !     call nextQ%enQueue(1_16)
 !     call beforeQ%enQueue(0_16) !dummy
 !     do while( .true. )
 !       if( nextQ%isEmpty() ) exit
 !       call solve( room,nextQ%deQueue(),beforeQ%deQueue() )
 !     end do
      
      
      
      
      if( count(flag(2:n)==-1) >=1 )then
        print*,'No';stop
      end if
      
      print*,'Yes'
      do i = 2,2
        print*,flag(i)
      end do
      
      
   !   print*,n,m
   !   do i = 1,m
   !     print*,a(i),b(i)
   !   end do
      
      
   !   do i= 1,n
   !     print*,'room',i,room(i)%numNei
   !     print*,room(i)%nei
   !   end do
      
      END PROGRAM