#include<bits/stdc++.h>
using namespace std;
int a[3],N,ans[100010];
char s[100010][3];
int main(){
	cin>>N>>a[0]>>a[1]>>a[2];
	for (int i=1; i<=N; i++) scanf("%s",s[i]);
	int flag = 1;
	for (int i=1; i<=N; i++){
		int x = s[i][0]-'A', y = s[i][1]-'A';
		if (!a[x] && !a[y]){
			flag = 0;
			break;
		}
		if (a[x] > a[y]) ans[i] = y, a[x]--, a[y]++;
		else if (a[x] < a[y]) ans[i] = x, a[x]++, a[y]--;
		else {
			int x2 = s[i+1][0] - 'A', y2 = s[i+1][1] - 'A';
			if (x == x2 || x == y2) ans[i] = x, a[x]++, a[y]--;
			else ans[i] = y, a[y]++, a[x]--;
		}
	}
	if (!flag) puts("No");
	else {
		puts("Yes");
		for (int i=1; i<=N; i++) printf("%c\n",ans[i]+'A');
	}
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 a[3],N,ans[100010];
char s[100010][3];
int main(){
	cin>>N>>a[0]>>a[1]>>a[2];
	for (int i=1; i<=N; i++) scanf("%s",s[i]);
	int flag = 1;
	for (int i=1; i<=N; i++){
		int x = s[i][0]-'A', y = s[i][1]-'A';
		if (!a[x] && !a[y]){
			flag = 0;
			break;
		}
		if (a[x] > a[y]) ans[i] = y, a[x]--, a[y]++;
		else if (a[x] < a[y]) ans[i] = x, a[x]++, a[y]--;
		else {
			int x2 = s[i+1][0] - 'A', y2 = s[i+1][1] - 'A';
			if (x == x2 || x == y2) ans[i] = x, a[x]++, a[y]--;
			else ans[i] = y, a[y]++, a[x]--;
		}
	}
	if (!flag) puts("No");
	else {
		puts("Yes");
		for (int i=1; i<=N; i++) printf("%c\n",ans[i]+'A');
	}
	return 0;
}
```