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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import itertools
input = sys.stdin.readline
from collections import defaultdict
from heapq import heappop, heappush

from decimal import *

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

##### 累乗 #####
# pow(x, y, z) -> x**y % z

##### 割り算の切り上げ #####
# tmp = -(-4 // 3)

##### dict の for文 #####
# for k, v in d.items():
#     print(k, v)

def inputInt(): return int(input())
def inputMap(): return map(int, input().split())
def inputList(): return list(map(int, input().split()))
def inputStr(): return input()[:-1]

inf = float('inf')
mod = 1000000007

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

def main():
	N,A,B,C = inputMap()
	stat = [0,0,0]
	statAB = [0,0,0]
	Q = []
	for i in range(N):
		s = inputStr()
		Q.append(s)
		if s == "AB":
			stat[0] += 1
			stat[1] += 1
			statAB[0] += 1
		elif s == "AC":
			stat[0] += 1
			stat[2] += 1
			statAB[1] += 1
		elif s == "BC":
			stat[1] += 1
			stat[2] += 1
			statAB[2] += 1

	delA = False
	delB = False
	delC = False
	if stat[0] <= A:
		delA = True
	if stat[1] <= B:
		delB = True
	if stat[2] <= C:
		delC = True

	targetA = False
	targetB = False
	targetC = False


	#print(stat)
	#print("{} {} {}".format(delA,delB,delC))

	hon = "Yes"
	ans = []
	for s in Q:
		#print(i)
		#print("{} {} {}".format(A,B,C))
		if s == "AB":
			statAB[0] -=1
			stat[0] -= 1
			stat[1] -= 1
			if delA == True:
				B += 1
				A -= 1
				ans.append("B")
			elif delB == True:
				B -= 1
				A += 1
				ans.append("A")
			elif targetA == True and B-1 >= 0:
				B -= 1
				A += 1
				ans.append("A")
			elif targetB == True and A-1 >= 0:
				B += 1
				A -= 1
				ans.append("B")
			elif statAB[0] < statAB[1]:
				B -= 1
				A += 1
				ans.append("A")
			elif A < B:
				B -= 1
				A += 1
				ans.append("A")
			else:
				B += 1
				A -= 1
				ans.append("B")
			if A < 0 or B < 0:
				hon = "No"
				break
		elif s == "AC":
			statAB[1] -=1
			stat[0] -= 1
			stat[2] -= 1
			if delA == True:
				C += 1
				A -= 1
				ans.append("C")
			elif delC == True:
				C -= 1
				A += 1
				ans.append("A")
			elif targetA == True and C-1 >= 0:
				C -= 1
				A += 1
				ans.append("A")
			elif targetC == True and A-1 >= 0:
				C += 1
				A -= 1
				ans.append("C")
			elif statAB[1] < statAB[2]:
				C += 1
				A -= 1
				ans.append("C")
			elif A < C:
				C -= 1
				A += 1
				ans.append("A")
			else:
				C += 1
				A -= 1
				ans.append("C")
			if A < 0 or C < 0:
				hon = "No"
				break
		elif s == "BC":
			statAB[2] -=1
			stat[1] -= 1
			stat[2] -= 1
			if delB == True:
				C += 1
				B -= 1
				ans.append("C")
			elif delC == True:
				C -= 1
				B += 1
				ans.append("B")
			elif targetB == True and C-1 >= 0:
				C -= 1
				B += 1
				ans.append("B")
			elif targetC == True and B-1 >= 0:
				C += 1
				B -= 1
				ans.append("C")
			elif statAB[0] < statAB[2]:
				C -= 1
				B += 1
				ans.append("B")
			elif B < C:
				C -= 1
				B += 1
				ans.append("B")
			else:
				C += 1
				B -= 1
				ans.append("C")
			if C < 0 or B < 0:
				hon = "No"
				break

		if stat[0] <= A:
			delA = True
		if stat[1] <= B:
			delB = True
		if stat[2] <= C:
			delC = True

	if hon == "No":
		print(hon)
	else:
		print(hon)
		for i in ans:
			print(i)


#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nCr mod m
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10**9+7):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res

# mを法とするaの乗法的逆元
def modinv(a, mod=10**9+7):
    return pow(a, mod-2, mod)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

# nHr mod m
# 問題によって、combination()を切り替えること
def H(n, r, mod=10**9+7):
	# comb = Combination(n+r-1, mod)
	# return comb(n+r-1, r)
    return combination(n+r-1, r, mod)


class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * modinv[self.mod%i] % self.mod
        return modinv

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# dfs のサンプル
def dfs(graph,parent,counter,edge):
    stk = []
    stk.append(edge)
    while len(stk) > 0:
        p = stk.pop()
        for e in graph[p]:
            if parent[p] == e:
                continue
            else:
                parent[e] = p
                counter[e] += counter[p]
                stk.append(e)

if __name__ == "__main__":
	main()