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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
string s[100000],m;
int main()
{
	long long n,a[3];
	cin>>n>>a[0]>>a[1]>>a[2];
	for(int t=0;t<n;t++)
	{
		cin>>s[t];
	}
	for(int t=0;t<n;t++)
	{
		if(a[s[t][0]-'A']<a[s[t][1]-'A'])
		{
			m+=s[t][0];
			a[s[t][0]-'A']++;
			a[s[t][1]-'A']--;
		}
		else if(a[s[t][0]-'A']==a[s[t][1]-'A'])
		{
			if(t!=n-1)
			{
		    	if(s[t][0]-'A'==s[t+1][0]-'A'||s[t][0]-'A'==s[t+1][1]-'A')
		    	{
		    		m+=s[t][0];
		    		a[s[t][0]-'A']++;
		    		a[s[t][1]-'A']--;
		    	}
		    	else
		    	{
		    		m+=s[t][1];
		    		a[s[t][1]-'A']++;
		    		a[s[t][0]-'A']--;
		    	}
		    }
		    else
		    {
		    	m+=s[t][0];
		    	a[s[t][0]-'A']++;
		    	a[s[t][1]-'A']--;
			}
	    }
		else
		{
			m+=s[t][1];
			a[s[t][1]-'A']++;
			a[s[t][0]-'A']--;
		}
		if(a[0]<0||a[1]<0||a[2]<0)
		{
			cout<<"No";
			return 0;
		}
	}
	cout<<"Yes"<<endl;
	for(int t=0;t<n;t++)
	{
		cout<<m[t]<<endl;
	}
	return 0;
}