#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define MP make_pair
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define PLI pair<ll, int>
#define PIL pair<int, ll>
#define STR string
#define Fst first
#define Snd second
#define SHN 1e9
#define LSHN 1e18
using namespace std;
int n;
int a[3];
string res="";
bool fl=1;
string s[100010];
bool b[100010][3];
int main()
{
	cin >>n;
	for (int i=0;i<3;i++) cin >>a[i];
	for (int i=0;i<n;i++)
		cin >>s[i],b[i][s[i][0]-'A']=1,b[i][s[i][1]-'A']=1;
	for (int i=0;i<n;i++)
	{
		char x=s[i][0],y=s[i][1];
		x-='A',y-='A';
		if (a[x]>a[y] ||
		(a[x]==a[y] && b[i+1][y] && !a[s[i+1][s[i+1][0]==y+'A' ? 1 : 0]-'A'])) swap(x,y);
		res+=(char)(x+'A');
		a[x]++,a[y]--; 
		if (a[y]<0) fl=0;
	}
	if (!fl) cout <<"No\n";
	else
	{
		cout <<"Yes\n";
		for (int i=0;i<n;i++) cout <<res[i]<<"\n";
	}
	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 <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define MP make_pair
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define PLI pair<ll, int>
#define PIL pair<int, ll>
#define STR string
#define Fst first
#define Snd second
#define SHN 1e9
#define LSHN 1e18
using namespace std;
int n;
int a[3];
string res="";
bool fl=1;
string s[100010];
bool b[100010][3];
int main()
{
	cin >>n;
	for (int i=0;i<3;i++) cin >>a[i];
	for (int i=0;i<n;i++)
		cin >>s[i],b[i][s[i][0]-'A']=1,b[i][s[i][1]-'A']=1;
	for (int i=0;i<n;i++)
	{
		char x=s[i][0],y=s[i][1];
		x-='A',y-='A';
		if (a[x]>a[y] ||
		(a[x]==a[y] && b[i+1][y] && !a[s[i+1][s[i+1][0]==y+'A' ? 1 : 0]-'A'])) swap(x,y);
		res+=(char)(x+'A');
		a[x]++,a[y]--; 
		if (a[y]<0) fl=0;
	}
	if (!fl) cout <<"No\n";
	else
	{
		cout <<"Yes\n";
		for (int i=0;i<n;i++) cout <<res[i]<<"\n";
	}
	return 0;
}
```