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<cstdlib>

#define int long long

const int MAXN=1e5;

int opt[MAXN+5],stp[MAXN+5];
char str[3];

inline bool compare(const char a[],const char b[]){
	if(a[0]==b[0] && a[1]==b[1])return true;
	return false;
}

int n;

inline void print_path(){
	puts("Yes");
	for(int i=1;i<=n;++i){
		if(stp[i]==1)puts("A");
		else if(stp[i]==2)puts("B");
		else puts("C");
	}
}

void dfs(const int i,const int a,const int b,const int c){
	if(a<0 || b<0 || c<0)return;
	if(i==n+1){
		print_path();
		exit(0);
	}
	if(opt[i]==1){
		stp[i]=1;
		dfs(i+1,a+1,b-1,c);
		stp[i]=2;
		dfs(i+1,a-1,b+1,c);
	}
	else if(opt[i]==2){
		stp[i]=1;
		dfs(i+1,a+1,b,c-1);
		stp[i]=3;
		dfs(i+1,a-1,b,c+1);
	}
	else{
		stp[i]=2;
		dfs(i+1,a,b+1,c-1);
		stp[i]=3;
		dfs(i+1,a,b-1,c+1);
	}
}

int a,b,c;

signed main(){
	scanf("%lld %lld %lld %lld",&n,&a,&b,&c);
	for(int i=1;i<=n;++i){
		scanf("%s",str);
		if(compare(str,"AB"))opt[i]=1;
		else if(compare(str,"AC"))opt[i]=2;
		else opt[i]=3;
	}
	dfs(1,a,b,c);
	puts("No");
	return 0;
}