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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
char s[10],ans[N],s1[10];
int main(){
	int n,a[3];
	scanf("%d%d%d%d%s",&n,a,a+1,a+2,s);
	if(a[s[0]-'A']+a[s[1]-'A']==0)return puts("No"),0;
	if(a[0]+a[1]+a[2]==1){
		ans[0]=a[s[0]-'A']==0?s[0]:s[1];
		swap(a[s[0]-'A'],a[s[1]-'A']);
		for(int i=1;i<n;i++){
			scanf("%s",s);
			if(a[s[0]-'A']+a[s[1]-'A']==0)return puts("No"),0;
			ans[i]=a[s[0]-'A']==0?s[0]:s[1];
			swap(a[s[0]-'A'],a[s[1]-'A']);
		}
		puts("Yes");
		for(int i=0;i<n;i++)printf("%c\n",ans[i]);
		return 0;
	}
	puts("Yes");
	for(int i=1;i<n;i++){
		int d;scanf("%s",s1);
		if(a[s[0]-'A']==a[s[1]-'A'])d=(s[1]==s1[0]||s[1]==s1[1]);
		else d=(a[s[0]-'A']>a[s[1]-'A']);
		a[s[d]-'A']++,a[s[!d]-'A']--;
		printf("%c\n",s[d]);
		s[0]=s1[0],s[1]=s1[1];
	}
	printf("%c\n",s[a[s[0]-'A']>0]);
	return 0;
}