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 ** 9)

n,a,b,c = list(map(int,input().split()))
s = []
for i in range(n):
  s.append(input())

ans = []
def dfs(i,a,b,c):
  if i == n:
    print('Yes')
    for ai in ans:
      print(ai)
    exit()
  si = s[i]
  if si == 'AB':
    if a == 0 and b == 0:
      return
    elif a == 0 and b > 0:
      ans.append('A')
      dfs(i+1,a+1,b-1,c)
    elif a > 0 and b == 0:
      ans.append('B')
      dfs(i+1,a-1,b+1,c)
    else:
      ans.append('A')
      dfs(i+1,a+1,b-1,c)
      ans.pop()
      ans.append('B')
      dfs(i+1,a-1,b+1,c)
  elif si == 'BC':
    if b == 0 and c == 0:
      return
    elif b == 0 and c > 0:
      ans.append('B')
      dfs(i+1,a,b+1,c-1)
    elif b > 0 and c == 0:
      ans.append('C')
      dfs(i+1,a,b-1,c+1)
    else:
      ans.append('B')
      dfs(i+1,a,b+1,c-1)
      ans.pop()
      ans.append('C')
      dfs(i+1,a,b-1,c+1)
  else:
    if c == 0 and a == 0:
      return
    elif c ==0 and a > 0:
      ans.append('C')
      dfs(i+1,a-1,b,c+1)
    elif c > 0 and a == 0:
      ans.append('A')
      dfs(i+1,a+1,b,c-1)
    else:
      ans.append('C')
      dfs(i+1,a-1,b,c+1)
      ans.pop()
      ans.append('A')
      dfs(i+1,a+1,b,c-1)

dfs(0,a,b,c)
print('No')