#include<bits/stdc++.h>
using namespace std;
const int N=200010;
int n,a[4],ans[N];
char s[N][4];
int main()
{
	cin>>n>>a[0]>>a[1]>>a[2];
	int flag=0;
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	for(int i=1;i<=n;i++)
	{
		int x=s[i][1]-'A',y=s[i][2]-'A';
		if(a[x]<a[y])
			a[x]++,a[y]--,ans[i]=x;
		else if(a[y]<a[x])
			a[x]--,a[y]++,ans[i]=y;
		else
		{
			if(a[x]==1&&!a[3-x-y])
			{
				if(i==n||(s[i+1][1]!=3-x-y+'A'&&s[i+1][2]!=3-x-y+'A'))
					a[x]--,a[y]++,ans[i]=y;
				else
				{
					int x1=s[i+1][1]-'A',x2=s[i+1][2]-'A';
					if(x1==x||x2==x)
						ans[i]=x,a[x]++,a[y]--;
					else
						ans[i]=y,a[x]--,a[y]++;
				}
			}
			else
				a[x]--,a[y]++,ans[i]=y;
		}
		if(a[x]<0||a[y]<0)
		{
			flag=1;
			break;
		}
	}
	if(flag)
		cout<<"No";
	else
	{
		cout<<"Yes"<<endl;
		for(int i=1;i<=n;i++)
			cout<<(char)(ans[i]+'A')<<endl;
	}
	return 0;
}

### Prompt

Please create a solution in CPP to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
const int N=200010;
int n,a[4],ans[N];
char s[N][4];
int main()
{
	cin>>n>>a[0]>>a[1]>>a[2];
	int flag=0;
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	for(int i=1;i<=n;i++)
	{
		int x=s[i][1]-'A',y=s[i][2]-'A';
		if(a[x]<a[y])
			a[x]++,a[y]--,ans[i]=x;
		else if(a[y]<a[x])
			a[x]--,a[y]++,ans[i]=y;
		else
		{
			if(a[x]==1&&!a[3-x-y])
			{
				if(i==n||(s[i+1][1]!=3-x-y+'A'&&s[i+1][2]!=3-x-y+'A'))
					a[x]--,a[y]++,ans[i]=y;
				else
				{
					int x1=s[i+1][1]-'A',x2=s[i+1][2]-'A';
					if(x1==x||x2==x)
						ans[i]=x,a[x]++,a[y]--;
					else
						ans[i]=y,a[x]--,a[y]++;
				}
			}
			else
				a[x]--,a[y]++,ans[i]=y;
		}
		if(a[x]<0||a[y]<0)
		{
			flag=1;
			break;
		}
	}
	if(flag)
		cout<<"No";
	else
	{
		cout<<"Yes"<<endl;
		for(int i=1;i<=n;i++)
			cout<<(char)(ans[i]+'A')<<endl;
	}
	return 0;
}

```