#include<bits/stdc++.h>
using namespace std;

const int N=1000005;
int n,a,b,c,d[N],ans[N];
char s[N];

int main()
{
	scanf("%d%d%d%d",&n,&a,&b,&c);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s);
		if(s[0]=='A'&&s[1]=='B') d[i]=1;
		else if(s[0]=='B') d[i]=2; else d[i]=3;
	}
	for(int i=1;i<=n;i++)
	{
		if(d[i]==1&&a+b==0||d[i]==2&&b+c==0||d[i]==3&&a+c==0)
		{
			printf("No\n");
			return 0;
		}
		if(d[i]==1&&a==1&&b==1&&d[i+1]>1)
		{
			if(d[i+1]==2) ans[i]=2,a--,b++;
			if(d[i+1]==3) ans[i]=1,a++,b--;
		}
		else if(d[i]==1&&b>a) ans[i]=1,a++,b--;
		else if(d[i]==1) ans[i]=2,a--,b++;
		if(d[i]==2&&b==1&&c==1&&d[i+1]%2)
		{
			if(d[i+1]==1) ans[i]=2,b++,c--;
			if(d[i+1]==3) ans[i]=3,b--,c++;
		}
		else if(d[i]==2&&b<c) ans[i]=2,b++,c--;
		else if(d[i]==2) ans[i]=3,b--,c++;
		if(d[i]==3&&a==1&&c==1&&d[i+1]<3&&d[i+1])
		{
			if(d[i+1]==1) ans[i]=1,a++,c--;
			if(d[i+1]==2) ans[i]=3,a--,c++;
		}
		else if(d[i]==3&&a<c) ans[i]=1,a++,c--;
		else if(d[i]==3) ans[i]=3,a--,c++;
	}
	printf("Yes\n");
	for(int i=1;i<=n;i++) printf("%c\n",(char('A'+ans[i]-1)));
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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<bits/stdc++.h>
using namespace std;

const int N=1000005;
int n,a,b,c,d[N],ans[N];
char s[N];

int main()
{
	scanf("%d%d%d%d",&n,&a,&b,&c);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s);
		if(s[0]=='A'&&s[1]=='B') d[i]=1;
		else if(s[0]=='B') d[i]=2; else d[i]=3;
	}
	for(int i=1;i<=n;i++)
	{
		if(d[i]==1&&a+b==0||d[i]==2&&b+c==0||d[i]==3&&a+c==0)
		{
			printf("No\n");
			return 0;
		}
		if(d[i]==1&&a==1&&b==1&&d[i+1]>1)
		{
			if(d[i+1]==2) ans[i]=2,a--,b++;
			if(d[i+1]==3) ans[i]=1,a++,b--;
		}
		else if(d[i]==1&&b>a) ans[i]=1,a++,b--;
		else if(d[i]==1) ans[i]=2,a--,b++;
		if(d[i]==2&&b==1&&c==1&&d[i+1]%2)
		{
			if(d[i+1]==1) ans[i]=2,b++,c--;
			if(d[i+1]==3) ans[i]=3,b--,c++;
		}
		else if(d[i]==2&&b<c) ans[i]=2,b++,c--;
		else if(d[i]==2) ans[i]=3,b--,c++;
		if(d[i]==3&&a==1&&c==1&&d[i+1]<3&&d[i+1])
		{
			if(d[i+1]==1) ans[i]=1,a++,c--;
			if(d[i+1]==2) ans[i]=3,a--,c++;
		}
		else if(d[i]==3&&a<c) ans[i]=1,a++,c--;
		else if(d[i]==3) ans[i]=3,a--,c++;
	}
	printf("Yes\n");
	for(int i=1;i<=n;i++) printf("%c\n",(char('A'+ans[i]-1)));
}
```