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:
from itertools import combinations

N, A, B, C = map(int, input().split())
S = [input() for _ in range(N)]

def sol():
    now = {'A': A, 'B': B, 'C': C}
    ret = []
    for i, s in enumerate(S):
        u, v = s[0], s[1]
        if now[u] == now[v] == 0:
            return []
        if now[u] > now[v]:
            ret.append(v)
            now[u] -= 1
            now[v] += 1
            continue
        if now[u] < now[v]:
            ret.append(u)
            now[u] += 1
            now[v] -= 1
            continue
        if i == N - 1:
            ret.append(u)
            continue

        nx = S[i + 1]
        if u in nx:
            ret.append(u)
            now[u] += 1
            now[v] -= 1
            continue
        if v in nx:
            ret.append(v)
            now[v] += 1
            now[u] -= 1
            continue

    return ret

ans = sol()
if len(ans) == 0:
    print('No')
    exit()

print('Yes')
print(*ans, sep='\n')