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>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define ll long long int

ll n,abc[4],a,b,c;
char let[4];
char s[100001][3];
char ans[100001];

void swap(int i,int p,int q){
  if(abc[p]>abc[q]){
      ans[i] = let[q];
      abc[p]--; abc[q]++;}
    else{
      ans[i] = let[p];
      abc[q]--; abc[p]++;}
  return;
}

int part2(){
  int p,q;
  for(ll i=0;i<n;i++){
    p = s[i][0]-'A';
    q = s[i][1]-'A';
    if(abc[p]==0 && abc[q]==0) return 0;
    swap(i,p,q);
  }
  return 1;
}

int part1(){
  int p,q;
  for(ll i=0;i<n;i++){
    p = s[i][0]-'A';
    q = s[i][1]-'A';
    if(abc[p]==0 && abc[q]==0) return 0;
    if(abc[p]==1 && abc[q]==1){
      int pnext = s[i+1][0]-'A';
      int qnext = s[i+1][1]-'A';
      if(p==pnext || p==qnext){
	ans[i] = let[p];
	abc[p]++; abc[q]--;}
      else{
	ans[i] = let[q];
	abc[q]++; abc[p]--;}}
    else{
      swap(i,p,q);}}
  return 1;
}

int main(){
  scanf("%lld%lld%lld%lld",&n,&abc[0],&abc[1],&abc[2]);
  a = abc[0];
  b = abc[1];
  c = abc[2];
  let[0] = 'A';
  let[1] = 'B';
  let[2] = 'C';
  for(int i=0;i<n;i++){
    scanf("%s",s[i]);}

  int kano=1;
  if(a+b+c==2) kano = part1();
  else kano = part2();

  if(kano == 1){
    printf("Yes\n");
    for(ll i=0;i<n;i++){
      printf("%c\n",ans[i]);}
  }
  else{
    printf("No\n");}

  return 0;
}