Problem: 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.

Examples

Input

2 1 3 0
AB
AC


Output

Yes
A
C


Input

3 1 0 0
AB
BC
AB


Output

No


Input

1 0 9 0
AC


Output

No


Input

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


Output

Yes
C
B
B
C
C
B
A
A

Solution (in PYTHON3):

Solution:
def main():
    N,A,B,C=map(int,input().split())
    d={'AB':(0,1),'AC':(0,2),'BC':(1,2)}
    s=[d[input()] for _ in range(N)]
    abc=[A,B,C]
    ABC='ABC'
    ans=[]
    for i in range(N):
        x,y=s[i]
        # 強制終了
        if abc[x] == abc[y] == 0:
            print('No')
            exit()
        if abc[x] == abc[y] == 1:
            # 最後の要素 or 次に今回選ぶ可能性があるものがあるかどうか
            if (i == N-1) or (x in s[i+1]):
                ans.append(ABC[x])
                abc[x]+=1
                abc[y]-=1
            else:
                ans.append(ABC[y])
                abc[x]-=1
                abc[y]+=1
        # 2数の和が1 or 3以上の場合は大きい方をインクリメント, 小さい方をデクリメントする.
        # 2数が同じ場合には 後半の文字に対応する番号を優先する.
        else:
            if abc[x] > abc[y]:
                ans.append(ABC[y])
                abc[x]-=1
                abc[y]+=1
            else:
                ans.append(ABC[x])
                abc[x]+=1
                abc[y]-=1
    print('Yes')
    for a in ans:
        print(a)

main()