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 <cstdio>
#include <algorithm>
#define maxn 100010
using namespace std;
char in[3];
int n,v[3],p[maxn],q[maxn],ans[maxn];
int main(){
	int i;
	scanf("%d",&n);
	for(i=0;i<3;i++)scanf("%d",&v[i]);//v[0]代表'A'的数量,v[1]代表'B'的数量,v[2]代表'C'的数量
	for(i=1;i<=n;i++)scanf("%s",in),p[i]=in[0]-'A',q[i]=in[1]-'A';
	for(i=1;i<=n;i++){
		if(v[p[i]]>v[q[i]])swap(p[i],q[i]);//让v[]小的在前
		else if(v[p[i]]==v[q[i]]){
			if(i+1<=n&&(q[i]==p[i+1]||q[i]==q[i+1]))swap(p[i],q[i]);//让接下来马上要用到的,在前
		}
		if(!v[q[i]])return 0*printf("No\n");
		v[p[i]]++,v[q[i]]--,ans[i]=p[i];
	}
	printf("Yes\n");
	for(i=1;i<=n;i++)printf("%c\n",'A'+ans[i]);
	return 0;
}