#include<bits/stdc++.h>
#define lint long long
#define st first
#define nd second

using namespace std;

vector<string> v;
string ans="";
int n,A,B,C;
set<vector<int>> vis;

bool run(int i){ 
	if(i == n) return true;
	if(vis.count({i,A,B,C})) return false;

	int &pri = v[i][0]=='A'?A:(v[i][0]=='B'?B:C);
	int &sec = v[i][1]=='A'?A:(v[i][1]=='B'?B:C);
 	
	if(pri - 1 >= 0){
		pri--;
		sec++;
		ans[i]=v[i][1];
		if(run(i+1))return true;
		ans[i]=' ';
		pri++;
		sec--;
	}
	
	if(sec - 1 >= 0){
		pri++;
		sec--;
		ans[i]=v[i][0];
		if(run(i+1))return true;
		ans[i]=' ';
		pri--;
		sec++;
	}
	vis.insert({i,A,B,C});
	return false;
}


//int t;
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	//cin>>t;
	//while(t--){}
	cin>>n>>A>>B>>C;
	v = vector<string> (n);
	for(string &s:v){
		ans += " ";
		cin>>s;
	}
	if(run(0)){
		cout<<"Yes\n";
		for(char &c:ans)
			cout<<c<<"\n";
	}else
		cout<<"No\n";


	return 0;	
}


### 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<bits/stdc++.h>
#define lint long long
#define st first
#define nd second

using namespace std;

vector<string> v;
string ans="";
int n,A,B,C;
set<vector<int>> vis;

bool run(int i){ 
	if(i == n) return true;
	if(vis.count({i,A,B,C})) return false;

	int &pri = v[i][0]=='A'?A:(v[i][0]=='B'?B:C);
	int &sec = v[i][1]=='A'?A:(v[i][1]=='B'?B:C);
 	
	if(pri - 1 >= 0){
		pri--;
		sec++;
		ans[i]=v[i][1];
		if(run(i+1))return true;
		ans[i]=' ';
		pri++;
		sec--;
	}
	
	if(sec - 1 >= 0){
		pri++;
		sec--;
		ans[i]=v[i][0];
		if(run(i+1))return true;
		ans[i]=' ';
		pri--;
		sec++;
	}
	vis.insert({i,A,B,C});
	return false;
}


//int t;
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	//cin>>t;
	//while(t--){}
	cin>>n>>A>>B>>C;
	v = vector<string> (n);
	for(string &s:v){
		ans += " ";
		cin>>s;
	}
	if(run(0)){
		cout<<"Yes\n";
		for(char &c:ans)
			cout<<c<<"\n";
	}else
		cout<<"No\n";


	return 0;	
}


```