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:
import sys

sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
f_inf = float('inf')
mod = 10 ** 9 + 7


def resolve():
    def calc(x, y):
        cnt[x] += 1
        cnt[y] -= 1
        res.append(S[x])

    n, *cnt = map(int, input().split())
    S = "ABC"
    query = []
    for _ in range(n):
        s = input().rstrip()
        a = 0 if s == "AB" else 0 if s == "AC" else 1
        b = 1 if s == "AB" else 2 if s == "AC" else 2
        query.append((a, b))
    query.append((None, None))

    res = []
    for i in range(n):
        a, b = query[i]
        if cnt[a] == cnt[b] == 0:
            print("No")
            return
        elif cnt[a] == 0:
            calc(a, b)
        elif cnt[b] == 0:
            calc(b, a)
        else:
            next_a, next_b = query[i + 1]
            if a == next_a or a == next_b:
                calc(a, b)
            else:
                calc(b, a)
    print("Yes")
    print(*res, sep="\n")


if __name__ == '__main__':
    resolve()