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 CPP):

Solution:
#include <stdio.h>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, n, a[3];
  scanf("%lld", &n);
  char s[10];
  ll p[n + 1], q[n + 1], ans[n];
  for(i = 0; i < 3; ++i) scanf("%lld", &a[i]);
  for(i = 0; i < n; ++i) {
    scanf("%s", s);
    if(s[0] == 'A') p[i] = 0;
    else p[i] = 1;
    if(s[1] == 'B') q[i] = 1;
    else q[i] = 2;
  }
  p[n] = q[n] = 0;
  for(i = 0; i < n; ++i) {
    if(!a[p[i]] && !a[q[i]]) {
      printf("No");
      return 0;
    }
    if(a[p[i]] == 1 && a[q[i]] == 1) {
      if(p[i] == p[i + 1] || p[i] == q[i + 1]) a[p[i]]++, a[q[i]]--, ans[i] = p[i];
      else a[p[i]]--, a[q[i]]++, ans[i] = q[i];
    } else {
      if(a[p[i]] > a[q[i]]) a[p[i]]--, a[q[i]]++, ans[i] = q[i];
      else a[p[i]]++, a[q[i]]--, ans[i] = p[i];
    }
  }
  printf("Yes\n");
  for(i = 0; i < n; ++i) printf("%c\n", ans[i] ? --ans[i] ? 'C' : 'B' : 'A');
  return 0;
}