#include <cstdio>
typedef long long ll;
inline int rd(){
    int x=0,p=1;
    char a=getchar();
    while((a<48||a>57)&&a!='-')a=getchar();
    if(a=='-')p=-p,a=getchar();
    while(a>47&&a<58)x=(x<<1)+(x<<3)+(a&15),a=getchar();
    return x*p;
}
const int N=200002;
int n;
int a[5];
char s[N][3],ans[N];
int main(){
	n=rd();
	for(int i=0;i<=2;i++)a[i]=rd();
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]);
	for(int i=1;i<=n;i++){
		int x=s[i][0]-'A',y=s[i][1]-'A';
		if(a[x]<a[y])a[x]++,a[y]--,ans[i]=x+'A';
		else if(a[x]>a[y])a[y]++,a[x]--,ans[i]=y+'A';
		else{
			int z=3-x-y;
			if(a[x]==1&&a[z]==0){
				if(i==n||(s[i+1][0]!=z+'A'&&s[i+1][1]!=z+'A'))a[y]++,a[x]--,ans[i]=y+'A';
				else{
					int xx=s[i+1][0]-'A',yy=s[i+1][1]-'A';
					if(x==xx||x==yy)a[x]++,a[y]--,ans[i]=x+'A';
					else a[y]++,a[x]--,ans[i]=y+'A';
				}
			}
			else a[y]++,a[x]--,ans[i]=y+'A';
		}
		if(a[x]<0||a[y]<0)return puts("No"),0;
	}
	puts("Yes");
	for(int i=1;i<=n;i++)printf("%c\n",ans[i]);
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#include <cstdio>
typedef long long ll;
inline int rd(){
    int x=0,p=1;
    char a=getchar();
    while((a<48||a>57)&&a!='-')a=getchar();
    if(a=='-')p=-p,a=getchar();
    while(a>47&&a<58)x=(x<<1)+(x<<3)+(a&15),a=getchar();
    return x*p;
}
const int N=200002;
int n;
int a[5];
char s[N][3],ans[N];
int main(){
	n=rd();
	for(int i=0;i<=2;i++)a[i]=rd();
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]);
	for(int i=1;i<=n;i++){
		int x=s[i][0]-'A',y=s[i][1]-'A';
		if(a[x]<a[y])a[x]++,a[y]--,ans[i]=x+'A';
		else if(a[x]>a[y])a[y]++,a[x]--,ans[i]=y+'A';
		else{
			int z=3-x-y;
			if(a[x]==1&&a[z]==0){
				if(i==n||(s[i+1][0]!=z+'A'&&s[i+1][1]!=z+'A'))a[y]++,a[x]--,ans[i]=y+'A';
				else{
					int xx=s[i+1][0]-'A',yy=s[i+1][1]-'A';
					if(x==xx||x==yy)a[x]++,a[y]--,ans[i]=x+'A';
					else a[y]++,a[x]--,ans[i]=y+'A';
				}
			}
			else a[y]++,a[x]--,ans[i]=y+'A';
		}
		if(a[x]<0||a[y]<0)return puts("No"),0;
	}
	puts("Yes");
	for(int i=1;i<=n;i++)printf("%c\n",ans[i]);
    return 0;
}
```