
#include <bits/stdc++.h>

#define o cout<<"Hm"<<'\n';
#define ll long long
#define pb push_back
#define mp make_pair
#define ld long double

using namespace std;

const int N = 2e5 + 100;
const int mod = 1e9 + 7;

int main ()
{
	#define int ll
	int n, a[3];
	cin >> n >> a[0] >> a[1] >> a[2];
	string ss[n + 1];
	string ans = "Yes\n";
	for (int i = 1; i <= n; i++)
	{
		cin >> ss[i];
	}
	for (int i = 1; i <= n; i++)
	{
		int x = ss[i][0] - 'A';
		int y = ss[i][1] - 'A';
		if (!a[x] && !a[y])
		{
			ans = "No";
			break;
		}
		if (!a[x])
		{
			a[x]++;
			a[y]--;
			ans += x + 'A';
		}
		else if (!a[y])
		{
			a[x]--;
			a[y]++;
			ans += y + 'A';
		}
		else if (a[x] == 1 && a[y] == 1 && i + 1 <= n)
		{
			if (ss[i + 1][0] - 'A' == x || ss[i + 1][1] - 'A' == x)
			{
				a[x]++;
				a[y]--;
				ans += x + 'A';
			}
			else {
				a[x]--;
				a[y]++;
				ans += y + 'A';
			}
		}
		else {
			if (a[x] > a[y])
			{
				a[x]--;
				a[y]++;
				ans += y + 'A';
			}
			else {
				a[x]++;
				a[y]--;
				ans += x + 'A';
			}
		}
		ans += '\n';
	}
	if (a[0] < 0 || a[1] < 0 || a[2] < 0)
	{
		ans = "No";
	}  
	cout << ans << '\n';
}



### Prompt

Your task is to create 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 o cout<<"Hm"<<'\n';
#define ll long long
#define pb push_back
#define mp make_pair
#define ld long double

using namespace std;

const int N = 2e5 + 100;
const int mod = 1e9 + 7;

int main ()
{
	#define int ll
	int n, a[3];
	cin >> n >> a[0] >> a[1] >> a[2];
	string ss[n + 1];
	string ans = "Yes\n";
	for (int i = 1; i <= n; i++)
	{
		cin >> ss[i];
	}
	for (int i = 1; i <= n; i++)
	{
		int x = ss[i][0] - 'A';
		int y = ss[i][1] - 'A';
		if (!a[x] && !a[y])
		{
			ans = "No";
			break;
		}
		if (!a[x])
		{
			a[x]++;
			a[y]--;
			ans += x + 'A';
		}
		else if (!a[y])
		{
			a[x]--;
			a[y]++;
			ans += y + 'A';
		}
		else if (a[x] == 1 && a[y] == 1 && i + 1 <= n)
		{
			if (ss[i + 1][0] - 'A' == x || ss[i + 1][1] - 'A' == x)
			{
				a[x]++;
				a[y]--;
				ans += x + 'A';
			}
			else {
				a[x]--;
				a[y]++;
				ans += y + 'A';
			}
		}
		else {
			if (a[x] > a[y])
			{
				a[x]--;
				a[y]++;
				ans += y + 'A';
			}
			else {
				a[x]++;
				a[y]--;
				ans += x + 'A';
			}
		}
		ans += '\n';
	}
	if (a[0] < 0 || a[1] < 0 || a[2] < 0)
	{
		ans = "No";
	}  
	cout << ans << '\n';
}



```