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: Accepted)

from collections import defaultdict as dd
from collections import deque
import bisect
import heapq

def ri():
    return int(input())

def rl():
    return list(map(int, input().split()))


def solve():
    n, a, b, c = rl()
    moves = []
    ans = []
    abc = [a,b,c]
    for i in range(n):
        moves.append(input())
    moves.append(moves[-1])

    for i in range(len(moves) - 1):
        curr = moves[i]
        nxt = moves[i + 1]
        x = {"A":0, "B":1, "C":2}[curr[0]]
        y = {"A":0, "B":1, "C":2}[curr[1]]
        nx = {"A":0, "B":1, "C":2}[nxt[0]]
        ny = {"A":0, "B":1, "C":2}[nxt[1]]

        if abc[x] == 0 and abc[y] == 0:
            print ("No")
            return
        elif abc[x] == 0:
            abc[x] += 1
            abc[y] -= 1
            ans.append(curr[0])
        elif abc[y] == 0:
            abc[x] -= 1
            abc[y] += 1
            ans.append(curr[1])
        elif x == y and nx == ny:
            abc[x] += 1
            abc[y] -= 1
            ans.append(curr[0])
        else:
            common = list(set([x,y]).intersection(set([nx,ny])))[0]
            if x == common:
                other = y
                ans.append(curr[0])
            else:
                other = x
                ans.append(curr[1])
            abc[common] += 1
            abc[other] -= 1

    print ("Yes")
    for l in ans:
        print (l)


mode = 's'

if mode == 'T':
    t = ri()
    for i in range(t):
        solve()
else:
    solve()