Problem:
Score : 600 points

Problem Statement

There is a game that involves three variables, denoted A, B, and C.

As the game progresses, there will be N events where you are asked to make a choice.
Each of these choices is represented by a string s_i. If s_i is AB, you must add 1 to A or B then subtract 1 from the other; if s_i is AC, you must add 1 to A or C then subtract 1 from the other; if s_i is BC, you must add 1 to B or C then subtract 1 from the other.

After each choice, none of A, B, and C should be negative.

Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.

Constraints

1 \leq N \leq 10^5

0 \leq A,B,C \leq 10^9

N, A, B, C are integers.

s_i is AB, AC, or BC.

Input

Input is given from Standard Input in the following format:

N A B C
s_1
s_2
:
s_N

Output

If it is possible to make N choices under the condition, print Yes; otherwise, print No.

Also, in the former case, show one such way to make the choices in the subsequent N lines. The (i+1)-th line should contain the name of the variable (A, B, or C) to which you add 1 in the i-th choice.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB

Sample Output 4

Yes
C
B
B
C
C
B
A
A

Solution in Fortran (GNU Fortran 9.2.1) (judge verdict: Accepted)

program three_variables_game
  implicit none
  integer :: n, r(4) = 0, x(100001) = 0, i, j, p(2) = 0
  character(1) :: c(4) = (/ "A", "B", "X", "C" /)
  character(2) :: s
  read(*,*) n, r(1), r(2), r(4)
  if (sum(r) == 0) then
    write(*,'(a)') "No"
    stop
  end if
  do i = 1, n
    read(*,*) s
    x(i) = hash(s)
  end do
  if (sum(r) == 1) then
    j = maxloc(r,1)
    call simulate(.false.)
    r = 0
    r(j) = 1
    call simulate(.true.)
    stop
  end if
  call get_index(x(1),p)
  if (r(p(1)) == 0 .and. r(p(2)) == 0) then
    write(*,'(a)') "No"
    stop
  else
    write(*,'(a)') "Yes"
  end if
  do i = 1, n
    call get_index(x(i),p)
    if (r(p(1)) == 0) then
      r(p(1)) = r(p(1))+1
      r(p(2)) = r(p(2))-1
      write(*,'(a)') c(p(1))
    else if (r(p(2)) == 0) then
      r(p(1)) = r(p(1))-1
      r(p(2)) = r(p(2))+1
      write(*,'(a)') c(p(2))
    else if (sum(r) == 2 .and. r(p(1))*r(p(2)) == 1 .and. i < n .and. x(i) /= x(i+1)) then
      j = 1
      if (p(2) == and(x(i),x(i+1))) j = 2
      r(p(j)) = r(p(j))+1
      r(p(3-j)) = r(p(3-j))-1
      write(*,'(a)') c(p(j))
    else if (r(p(1)) > r(p(2))) then
      r(p(1)) = r(p(1))-1
      r(p(2)) = r(p(2))+1
      write(*,'(a)') c(p(2))
    else
      r(p(1)) = r(p(1))+1
      r(p(2)) = r(p(2))-1
      write(*,'(a)') c(p(1))
    end if
  end do
contains
  integer function hash(s)
    character(2), intent(in) :: s
    select case (s)
      case ("AB")
        hash = 3
      case ("AC")
        hash = 5
      case ("BC")
        hash = 6
    end select
  end
  subroutine get_index(x,p)
    integer, intent(in) :: x
    integer, intent(out) :: p(2)
    integer :: i, k
    i = 0
    do k = 0, 2
      if (.not.btest(x,k)) cycle
      i = i+1
      p(i) = lshift(1,k)
    end do
  end
  subroutine simulate(out)
    logical, intent(in) :: out
    if (out) write(*,'(a)') "Yes"
    do i = 1, n
      call get_index(x(i),p)
      if (r(p(1)) /= 0) then
        r(p(1)) = r(p(1))-1
        r(p(2)) = r(p(2))+1
        if (out) write(*,'(a)') c(p(2))
      else if (r(p(2)) /= 0) then
        r(p(1)) = r(p(1))+1
        r(p(2)) = r(p(2))-1
        if (out) write(*,'(a)') c(p(1))
      else
        write(*,'(a)') "No"
        stop
      end if
    end do
  end
end program three_variables_game