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 x[100010][2];
int ans[100010];
int n, a[3];
void f(int i,int p,int q){
	ans[i] = p;
	a[p]++;
	a[q]--;
}
int main(){
	scanf("%d%d%d%d",&n,a,a+1,a+2);
	for(int i = 0; i < n; i++){
		char s[3];
		scanf("%s",s);
		x[i][0] = s[0] - 'A';
		x[i][1] = s[1] - 'A';
	}
	for(int i = 0; i < n; i++){
		int p = x[i][0], q = x[i][1];
		if(!a[p]&&!a[q]){
			puts("No");
			return 0;
		}
		else if(!a[p])
            f(i,p,q);
		else if(!a[q])
            f(i,q,p);
		else{
			if(i != n - 1 && p == 3 - x[i + 1][0] - x[i + 1][1])
                f(i,q,p);
			else f(i,p,q);
		}
	}
	puts("Yes");
	for(int i = 0; i < n; i++)
        printf("%c\n",ans[i]+'A');
    return 0;
}