#include<bits/stdc++.h>
#define mod 1000000007
#define ll long long int
#define ar array
#define pb push_back
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int n,a,b,c;
	cin>>n>>a>>b>>c;
	map<char,int>m;
	m['A']= a;
	m['B'] = b;
	m['C'] = c;
	vector<char>v;
	vector<string>v2(n);
	int flag = 1;
	string s;
	for(int i = 0;i<n; i++)
		cin>>v2[i];
	for(int i = 0; i<n; i++)
	{
		s = v2[i];
		if(m[s[0]]>m[s[1]])
		{
			m[s[0]]--;
			if(m['A']<0 || m['B']<0 || m['C']<0) flag = 0;
			m[s[1]]++;
			v.pb(s[1]);
		}
		else
		{
			string t="";
			if(m[s[0]]==m[s[1]] && m[s[0]]==1)
				if(i!=n-1)
					t = v2[i+1];
			if(t.size()==0 || t[0]==s[0] || t[1]==s[0])
			{
				m[s[1]]--;
				if(m['A']<0 || m['B']<0 || m['C']<0) flag = 0;
				m[s[0]]++;
				v.pb(s[0]);
			}
			else
			{
				m[s[0]]--;
				if(m['A']<0 || m['B']<0 || m['C']<0) flag = 0;
				m[s[1]]++;
				v.pb(s[1]);
			}
		}
	}
	if(flag)
	{
		cout<<"Yes"<<endl;
		for(char i:v)
			cout<<i<<endl;
	}
	else
		cout<<"No"<<endl;
	return 0;
}
### Prompt

Your challenge is to write a CPP solution 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>
#define mod 1000000007
#define ll long long int
#define ar array
#define pb push_back
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int n,a,b,c;
	cin>>n>>a>>b>>c;
	map<char,int>m;
	m['A']= a;
	m['B'] = b;
	m['C'] = c;
	vector<char>v;
	vector<string>v2(n);
	int flag = 1;
	string s;
	for(int i = 0;i<n; i++)
		cin>>v2[i];
	for(int i = 0; i<n; i++)
	{
		s = v2[i];
		if(m[s[0]]>m[s[1]])
		{
			m[s[0]]--;
			if(m['A']<0 || m['B']<0 || m['C']<0) flag = 0;
			m[s[1]]++;
			v.pb(s[1]);
		}
		else
		{
			string t="";
			if(m[s[0]]==m[s[1]] && m[s[0]]==1)
				if(i!=n-1)
					t = v2[i+1];
			if(t.size()==0 || t[0]==s[0] || t[1]==s[0])
			{
				m[s[1]]--;
				if(m['A']<0 || m['B']<0 || m['C']<0) flag = 0;
				m[s[0]]++;
				v.pb(s[0]);
			}
			else
			{
				m[s[0]]--;
				if(m['A']<0 || m['B']<0 || m['C']<0) flag = 0;
				m[s[1]]++;
				v.pb(s[1]);
			}
		}
	}
	if(flag)
	{
		cout<<"Yes"<<endl;
		for(char i:v)
			cout<<i<<endl;
	}
	else
		cout<<"No"<<endl;
	return 0;
}
```