#include<cmath>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;

int n,a[3];
vector<char> ans;

int main(){
    cin>>n>>a[0]>>a[1]>>a[2];
    vector<string> s(n);
    for(int i=0;i<n;i++)cin>>s[i];
    
    for(int i=0;i<n;i++)
    {
    	int x=s[i][0]-'A',y=s[i][1]-'A',z=3-x-y;
	
	if(a[x]==0 && a[y]==0)
	{
		cout<<"No"<<"\n";
		return 0;
	}
    
    	else if(a[x]==1 && a[y]==1 && a[z]==0)
	{
		if(i==n-1)
		{
			a[x]++,a[y]--;
			ans.push_back(s[i][0]);
		}
		else if(s[i+1].find(s[i][0])!=string::npos)
		{
			a[x]++,a[y]--;		
		        ans.push_back(s[i][0]);
		}
		else
		{
			a[x]--,a[y]++;
			ans.push_back(s[i][1]);
		}
	}
	else if(a[x]<=a[y])
	{
		a[x]++,a[y]--;
		ans.push_back(s[i][0]);	
	}
	else if(a[x]>a[y])
	{
		a[x]--,a[y]++;
		ans.push_back(s[i][1]);	
	}
    }
    cout<<"Yes"<<"\n";
    for(int i=0;i<n;i++)
    {
    	cout<<ans[i]<<"\n";
    }
}

### Prompt

Create a solution in CPP for 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<cmath>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;

int n,a[3];
vector<char> ans;

int main(){
    cin>>n>>a[0]>>a[1]>>a[2];
    vector<string> s(n);
    for(int i=0;i<n;i++)cin>>s[i];
    
    for(int i=0;i<n;i++)
    {
    	int x=s[i][0]-'A',y=s[i][1]-'A',z=3-x-y;
	
	if(a[x]==0 && a[y]==0)
	{
		cout<<"No"<<"\n";
		return 0;
	}
    
    	else if(a[x]==1 && a[y]==1 && a[z]==0)
	{
		if(i==n-1)
		{
			a[x]++,a[y]--;
			ans.push_back(s[i][0]);
		}
		else if(s[i+1].find(s[i][0])!=string::npos)
		{
			a[x]++,a[y]--;		
		        ans.push_back(s[i][0]);
		}
		else
		{
			a[x]--,a[y]++;
			ans.push_back(s[i][1]);
		}
	}
	else if(a[x]<=a[y])
	{
		a[x]++,a[y]--;
		ans.push_back(s[i][0]);	
	}
	else if(a[x]>a[y])
	{
		a[x]--,a[y]++;
		ans.push_back(s[i][1]);	
	}
    }
    cout<<"Yes"<<"\n";
    for(int i=0;i<n;i++)
    {
    	cout<<ans[i]<<"\n";
    }
}

```