#include<bits/stdc++.h>
using namespace std;
#define ll long long int;
int h[100002],n,m,a[3],ans[100005],l,r,la,ra;
char s[100005][5];
int main(){
cin>>n;
for(int i=0;i<3;i++) cin>>a[i];
for(int i=0;i<n;i++) cin>>s[i];
for(int i=0;i<n;i++){
	l=s[i][0]-'A';
	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){
		la=s[i+1][0]-'A';
		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++){
	cout<<(char)ans[i]<<endl;
}
}

### Prompt

Develop a solution in CPP 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;
#define ll long long int;
int h[100002],n,m,a[3],ans[100005],l,r,la,ra;
char s[100005][5];
int main(){
cin>>n;
for(int i=0;i<3;i++) cin>>a[i];
for(int i=0;i<n;i++) cin>>s[i];
for(int i=0;i<n;i++){
	l=s[i][0]-'A';
	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){
		la=s[i+1][0]-'A';
		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++){
	cout<<(char)ans[i]<<endl;
}
}

```