#include<bits/stdc++.h>
using namespace std;
vector<char>v;
int main()
{
	int g[3];
	int n;
	cin>>n>>g[0]>>g[1]>>g[2];
	char c1,c2,c3,c4;
	cin>>c1>>c2;
	for(int i=0;i<n-1;i++)
	{
		cin>>c3>>c4;
		if(g[c1-'A']>g[c2-'A'])
			g[c1-'A']--,g[c2-'A']++,v.push_back(c2);
		else if(g[c1-'A']<g[c2-'A'])
			g[c1-'A']++,g[c2-'A']--,v.push_back(c1);
		else
		{
			if(c1==c3&&c2!=c4||c1==c4&&c2!=c3)
				g[c1-'A']++,g[c2-'A']--,v.push_back(c1);
			else
				g[c1-'A']--,g[c2-'A']++,v.push_back(c2);
		}
		if(g[c1-'A']<0||g[c2-'A']<0)
		{
			cout<<"No";
			return 0;
		}
		c1=c3;c2=c4;
	}
	if(g[c1-'A']>g[c2-'A'])
			g[c1-'A']--,g[c2-'A']++,v.push_back(c2);
		else 
			g[c1-'A']++,g[c2-'A']--,v.push_back(c1);
	if(g[c1-'A']<0||g[c2-'A']<0)
		{
			cout<<"No";
			return 0;
		}
//	cout<<g[0]<<g[1]<<g[2];
	cout<<"Yes"<<endl;
	for(int i=0;i<v.size();i++)
		cout<<v[i]<<endl;
	return 0;
}
### Prompt

Develop a solution in Cpp 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;
vector<char>v;
int main()
{
	int g[3];
	int n;
	cin>>n>>g[0]>>g[1]>>g[2];
	char c1,c2,c3,c4;
	cin>>c1>>c2;
	for(int i=0;i<n-1;i++)
	{
		cin>>c3>>c4;
		if(g[c1-'A']>g[c2-'A'])
			g[c1-'A']--,g[c2-'A']++,v.push_back(c2);
		else if(g[c1-'A']<g[c2-'A'])
			g[c1-'A']++,g[c2-'A']--,v.push_back(c1);
		else
		{
			if(c1==c3&&c2!=c4||c1==c4&&c2!=c3)
				g[c1-'A']++,g[c2-'A']--,v.push_back(c1);
			else
				g[c1-'A']--,g[c2-'A']++,v.push_back(c2);
		}
		if(g[c1-'A']<0||g[c2-'A']<0)
		{
			cout<<"No";
			return 0;
		}
		c1=c3;c2=c4;
	}
	if(g[c1-'A']>g[c2-'A'])
			g[c1-'A']--,g[c2-'A']++,v.push_back(c2);
		else 
			g[c1-'A']++,g[c2-'A']--,v.push_back(c1);
	if(g[c1-'A']<0||g[c2-'A']<0)
		{
			cout<<"No";
			return 0;
		}
//	cout<<g[0]<<g[1]<<g[2];
	cout<<"Yes"<<endl;
	for(int i=0;i<v.size();i++)
		cout<<v[i]<<endl;
	return 0;
}
```