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 Julia (1.4.0) (judge verdict: Runtime Error)

parseInt(x) = parse(Int,x)
function tie(s)#k[1]->"A",k[2]->"B",k[3]->"C"
    if s == "A"
        return 1
    end
    if s == "B"
        return 2
    end
    if s == "C"
        return 3
    end
end

function check(s1,s2)
    d=Dict()
    d[s1]=1
    d[s2]=2
    if !haskey(d,"A")
        return "A"
    end
    if !haskey(d,"B")
        return "B"
    end
    if !haskey(d,"C")
        return "C"
    end
end

function f()
    k=zeros(Int,3)#k[1]=a,k[2]=b,k[3]=c
    n,k[1],k[2],k[3]=map(x->parseInt(x),split(readline()))
    s=Matrix{SubString{String}}(undef,n,2)
    q=Vector{SubString{String}}(undef,n)
    for i=1:n
        s[i,1],s[i,2]=split.(readline(),"")
    end
    for i=1:n
        if k[1]+k[2]+k[3] == 0 #a+b+c=0
            print("No")
            exit()
        end
        if  k[tie(s[i,1])] == k[tie(s[i,2])] == 0 #両方0
            print("No")
            exit()
        end
        if k[tie(s[i,1])] == 0 #片方0-left
            q[i]=s[i,1]
            k[tie(s[i,1])] += 1
            k[tie(s[i,2])] -= 1
        elseif k[tie(s[i,2])] == 0 #片方0-right
            q[i]=s[i,2]
            k[tie(s[i,1])] -= 1
            k[tie(s[i,2])] += 1
        #両方1以上
        elseif k[1]+k[2]+k[3] == 2 && k[tie(s[i,1])] == k[tie(s[i,2])] ==1
            if i != n
                if s[i,1] != check(s[i+1,1],s[i+1,2])
                    q[i]=s[i,1]
                    k[tie(s[i,1])] += 1
                    k[tie(s[i,2])] -= 1
                else
                    q[i]=s[i,2]
                    k[tie(s[i,1])] -= 1
                    k[tie(s[i,2])] += 1
                end
            end
        else
            q[i]=s[i,1]
            k[tie(s[i,1])] += 1
            k[tie(s[i,2])] -= 1
        end    
    end
    println("Yes")
    for i=1:n
        println(q[i])
    end
end
    
f()