Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB

Sample Output 4

Yes
C
B
B
C
C
B
A
A

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void) {

  long n,a,b,c;
  scanf("%ld %ld %ld %ld", &n, &a, &b, &c);
  char s[n][3];
  for (long i = 0; i < n; i++) {
    scanf("%s", s[i]);
  }
  char operation[n];
  for (long i = 0; i < n; i++) {
    if (s[i][0] == 'A' && s[i][1] == 'B') {
      if (a == 0 && b == 0) {
        printf("No\n");
        return 0;
      } else if (b == 0) {
        a--;
        b++;
        operation[i] = 'B';
      } else if (a == 0) {
        a++;
        b--;
        operation[i] = 'A';
      } else if (i != n-1 && s[i+1][0] == 'B' && s[i+1][1] == 'C') {
        a--;
        b++;
        operation[i] = 'B';
      } else {
        a++;
        b--;
        operation[i] = 'A';
      }
    } else if (s[i][0] == 'A' && s[i][1] == 'C') {
      if (a == 0 && c == 0) {
        printf("No\n");
        return 0;
      } else if (c == 0) {
        a--;
        c++;
        operation[i] = 'C';
      } else if (a == 0) {
        a++;
        c--;
        operation[i] = 'A';
      } else if (i != n-1 && s[i+1][0] == 'B' && s[i+1][1] == 'C') {
        a--;
        c++;
        operation[i] = 'C';
      } else {
        a++;
        c--;
        operation[i] = 'A';
      }
    } else {
      if (b == 0 && c == 0) {
        printf("No\n");
        return 0;
      } else if (c == 0) {
        b--;
        c++;
        operation[i] = 'C';
      } else if (b == 0) {
        b++;
        c--;
        operation[i] = 'B';
      } else if (i != n-1 && s[i+1][0] == 'A' && s[i+1][1] == 'C') {
        b--;
        c++;
        operation[i] = 'C';
      } else {
        b++;
        c--;
        operation[i] = 'B';
      }
    }
  }
  printf("Yes\n");
  for (long i = 0; i < n; i++) {
    printf("%c\n", operation[i]);
  }

  return 0;
}