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
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
n, *abc = map(int, input().split())
s = [input().rstrip()for _ in range(n)]
cond = {'AB': (0, 1), 'AC': (0, 2), 'BC': (1, 2)}
legend = ('A', 'B', 'C')
ans = []


def dfs(i):
    if -1 in abc:
        return 0
    elif i == n:
        print('Yes')
        print(*ans, sep='\n')
        exit()
        # return 0
    else:
        idx1, idx2 = cond[s[i]]
        abc[idx1] += 1
        abc[idx2] -= 1
        ans.append(legend[idx1])
        dfs(i+1)
        ans.pop()
        abc[idx1] -= 1
        abc[idx2] += 1

        abc[idx1] -= 1
        abc[idx2] += 1
        ans.append(legend[idx2])
        dfs(i+1)
        ans.pop()
        abc[idx1] += 1
        abc[idx2] -= 1


dfs(0)
print('No')