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>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>

inline void read(int &T) {
	int x=0;bool f=0;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=!f;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	T=f?-x:x;
}

int n,a,b,c;
char ans[100001];
std::string cmd[100001];

void dfs(int step,int a,int b,int c) {
	if(step>n) {
		puts("Yes");
		for(int i=1;i<=n;++i) {
			printf("%c\n",ans[i]);
		}
		exit(0);
	}
	if(cmd[step]=="AB") {
		if(a) {
			ans[step]='B';
			dfs(step+1,a-1,b+1,c);
		}
		if(b) {
			ans[step]='A';
			dfs(step+1,a+1,b-1,c);
		}
	}
	if(cmd[step]=="AC") {
		if(a) {
			ans[step]='C';
			dfs(step+1,a-1,b,c+1);
		}
		if(c) {
			ans[step]='A';
			dfs(step+1,a+1,b,c-1);
		}
	}
	if(cmd[step]=="BC") {
		if(b) {
			ans[step]='C';
			dfs(step+1,a,b-1,c+1);
		}
		if(c) {
			ans[step]='B';
			dfs(step+1,a,b+1,c-1);
		}
	}
}

int main() {
	read(n),read(a),read(b),read(c);
	for(int i=1;i<=n;++i) std::cin>>cmd[i];
	dfs(1,a,b,c);puts("No");
	return 0;
}