#include <iostream>

using namespace std;

int main(){
	
	cin.tie(0); ios_base::sync_with_stdio(0);
	int n,en[3];
	cin>>n;
	string s[n];
	for(int i=0;i<3;i++){
		cin>>en[i];
	}
	for(int i=0;i<n;i++){
		cin>>s[i];
	}
	int la[n+1][3];
	la[n][0]=la[n][1]=la[n][2]=n;
	char res[n];
	for(int i=n-1;i>=0;i--){
		for(int j = 0;j<3;j++)la[i][j] = la[i+1][j];
		la[i][s[i][0]-'A']=i;
		la[i][s[i][1]-'A']=i;
	}
	for(int i=0;i<n;i++){
		int i1 = s[i][0]-'A';
		int i2 = s[i][1]-'A';
		int i3;
		for(i3=0;i3==i1||i3==i2;i3++);
		if(en[i1]==0&&en[i2]==0){
			cout<<"No\n";
			return 0;
		}
		if(en[i1]==1&&en[i2]==1){
			if(la[i][i3]<n){
				int wa;
				if(s[la[i][i3]][0]=='A'+i3)wa = s[la[i][i3]][1] - 'A';
				else wa = s[la[i][i3]][0] - 'A';
				res[i] = 'A'+wa;
				en[wa]++;
				if(i1!=wa)en[i1]--;
				else en[i2]--;
			}
			else {
				res[i] = 'A'+i2;
				en[i2]++;
				en[i1]--;
			}
		}
		else if(en[i1]>en[i2]){
			res[i] = 'A'+i2;
			en[i2]++;
			en[i1]--;
		}
		else{
			res[i] = 'A'+i1;
			en[i1]++;
			en[i2]--;
		}
	}
	cout<<"Yes\n";
	for(int i=0;i<n;i++)cout<<res[i]<<endl;
	
	return 0;
}
### Prompt

In cpp, your task is to solve 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 <iostream>

using namespace std;

int main(){
	
	cin.tie(0); ios_base::sync_with_stdio(0);
	int n,en[3];
	cin>>n;
	string s[n];
	for(int i=0;i<3;i++){
		cin>>en[i];
	}
	for(int i=0;i<n;i++){
		cin>>s[i];
	}
	int la[n+1][3];
	la[n][0]=la[n][1]=la[n][2]=n;
	char res[n];
	for(int i=n-1;i>=0;i--){
		for(int j = 0;j<3;j++)la[i][j] = la[i+1][j];
		la[i][s[i][0]-'A']=i;
		la[i][s[i][1]-'A']=i;
	}
	for(int i=0;i<n;i++){
		int i1 = s[i][0]-'A';
		int i2 = s[i][1]-'A';
		int i3;
		for(i3=0;i3==i1||i3==i2;i3++);
		if(en[i1]==0&&en[i2]==0){
			cout<<"No\n";
			return 0;
		}
		if(en[i1]==1&&en[i2]==1){
			if(la[i][i3]<n){
				int wa;
				if(s[la[i][i3]][0]=='A'+i3)wa = s[la[i][i3]][1] - 'A';
				else wa = s[la[i][i3]][0] - 'A';
				res[i] = 'A'+wa;
				en[wa]++;
				if(i1!=wa)en[i1]--;
				else en[i2]--;
			}
			else {
				res[i] = 'A'+i2;
				en[i2]++;
				en[i1]--;
			}
		}
		else if(en[i1]>en[i2]){
			res[i] = 'A'+i2;
			en[i2]++;
			en[i1]--;
		}
		else{
			res[i] = 'A'+i1;
			en[i1]++;
			en[i2]--;
		}
	}
	cout<<"Yes\n";
	for(int i=0;i<n;i++)cout<<res[i]<<endl;
	
	return 0;
}
```