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<cstdio>
#include<map>
using namespace std;
typedef long long int ll;

int h[100000];


int n;
int m;


int main(){
	int n,a[3];
	int ans[100005];
	char s[100005][5];
	scanf("%d",&n);
	for(int i=0;i<3;i++) scanf("%d",a+i);
	for(int i=0;i<n;i++) scanf("%s",s[i]);
	for(int i=0;i<n;i++){
		int l = s[i][0] - 'A';
		int r = s[i][1] - 'A';
		if(a[l]==0&&a[r]==0){
			printf("No\n");
			return 0;
		}
		if(a[l]>a[r]){
			a[r]++;
			a[l]--;
			ans[i] = 'A'+r;
		}else if(a[0]+a[1]+a[2]==2&&a[l]==1&&a[r]==1&&i<n-1){
			int la = s[1+i][0] - 'A';
			int ra = s[i+1][1] - 'A';
			if(la==l||ra==l){
			a[l]++;
			a[r]--;
			ans[i] = 'A'+l;
			}else{
			ans[i] = 'A'+r;
			a[l]--;
			a[r]++;
			}
		}else{
			a[l]++;
			a[r]--;
			ans[i] = 'A'+l;
		}
	}
	printf("Yes\n");
	for(int i=0;i<n;i++){
		printf("%c\n",(char)ans[i]);
	}
}