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 Kotlin (1.3.71) (judge verdict: Accepted)



import java.util.*

fun main() {
    val (n,a,b,c) = readLine()!!.split(" ").map { it.toInt() }
    val queries = Array(n){""}
    val sets = Array(3){ TreeSet<Int>() }
    val arr = IntArray(3)
    val m = charArrayOf('A','B','C')
    val ans = CharArray(n)

    arr[0] = a
    arr[1] = b
    arr[2] = c

    for(i in 0 until n) {
        queries[i] = readLine()!!
        for(v in queries[i]) {
            val d = v - 'A'
            sets[d].add(i)
        }
    }

    for(i in 0 until n) {
        val q = queries[i]
        val l = q[0] - 'A'
        val r = q[1] - 'A'
        var select = -1

        if(arr[l] > arr[r]) {
            arr[l]--
            arr[r]++
            select = r
        } else if(arr[l]<arr[r]) {
            arr[l]++
            arr[r]--
            select = l
        } else  {
            val nl = sets[l].higher(i) ?: Int.MAX_VALUE
            val nr = sets[r].higher(i) ?: Int.MAX_VALUE
            if(nl < nr) {
                arr[r]--
                arr[l]++
                select = l
            } else {
                arr[r]++
                arr[l]--
                select = r
            }
        }
        if(arr[l] < 0 || arr[r] < 0) {
            print("No")
            return
        }
        ans[i] = m[select]
    }
    println("Yes")
    for(i in 0 until n) {
        println(ans[i])
    }
    // 서로 갯수가 같으면
    // 다음번에 나올 횟수가 가장 적은 녀석으로?


}