#include <bits/stdc++.h>
using namespace std;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, a[3];  cin >> n >> a[0] >> a[1] >> a[2];
	vector<string> q(n);
	for(int i=0;i<n;i++)cin>>q[i];
	vector<char> result;
	for (int i = 0; i < n; ++i)
	{
		string s = q[i];
		int x = s[0] - 'A';
		int y = s[1] - 'A';
		if (!a[x] && !a[y])
		{      
			result.clear();
			break;
		}
	    if (a[x] > a[y])
		{
			--a[x];++a[y];
			result.push_back(s[1]);
			continue;
		}
	    if (a[x] < a[y])
		{
			++a[x]; --a[y];
			result.push_back(s[0]);
			continue;
		}
	    if (i + 1 < n && q[i + 1].find(s[0]) != string::npos)
		{
			++a[x]; --a[y];
			result.push_back(s[0]);
			continue;
		}
	    --a[x]; ++a[y];
		result.push_back(s[1]);
	}
	if (result.empty())
	{
		cout<<"No\n";
		return 0;
	}
	cout << "Yes\n"; 
	for(int i=0;i<result.size();i++)cout<<result[i]<<endl;
}
### Prompt

Please formulate 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>
using namespace std;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, a[3];  cin >> n >> a[0] >> a[1] >> a[2];
	vector<string> q(n);
	for(int i=0;i<n;i++)cin>>q[i];
	vector<char> result;
	for (int i = 0; i < n; ++i)
	{
		string s = q[i];
		int x = s[0] - 'A';
		int y = s[1] - 'A';
		if (!a[x] && !a[y])
		{      
			result.clear();
			break;
		}
	    if (a[x] > a[y])
		{
			--a[x];++a[y];
			result.push_back(s[1]);
			continue;
		}
	    if (a[x] < a[y])
		{
			++a[x]; --a[y];
			result.push_back(s[0]);
			continue;
		}
	    if (i + 1 < n && q[i + 1].find(s[0]) != string::npos)
		{
			++a[x]; --a[y];
			result.push_back(s[0]);
			continue;
		}
	    --a[x]; ++a[y];
		result.push_back(s[1]);
	}
	if (result.empty())
	{
		cout<<"No\n";
		return 0;
	}
	cout << "Yes\n"; 
	for(int i=0;i<result.size();i++)cout<<result[i]<<endl;
}
```