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<iostream>
#include<cstdio>
using namespace std;
int record[100005][2];
int res[100005];
int n, a[3];
int main(){
	scanf("%d%d%d%d",&n,&a[0],&a[1],&a[2]);
	for(int i=0;i<n;i++){
		char s[3];
		scanf("%s",s);
		record[i][0]=s[0]-'A';
		record[i][1]=s[1]-'A';
	}

	for(int i=0;i<n;i++){
		int op1=record[i][0], op2=record[i][1];
		if(!a[op1]&&!a[op2]){
		        printf("No\n");
			return 0;
		}
		else if(!a[op1]) {
                       	res[i]=op1;
	                a[op1]++;
	                a[op2]--;
                    }
		else if(!a[op2]){
                      	res[i]=op2;
	                 a[op2]++;
	                a[op1]--;
                 }
		else{
			if(i!=n-1&&op1==3-record[i+1][0]-record[i+1][1]){
                      	res[i]=op2;
	                a[op2]++;
	                a[op1]--;
                          }
			else  {
                       	res[i]=op1;
	                a[op1]++;
	                a[op2]--;
                    }
		}
	}
	printf("Yes\n");
	for(int i=0;i<n;i++)
              printf("%c\n",res[i]+'A');
}