Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB

Sample Output 4

Yes
C
B
B
C
C
B
A
A

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<cmath>
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<map>
#include<queue>
#include<set>
#include<vector>
#include<bitset>
#define int long long
#define lowbit(x) ((x)&(-x))
#define hh putchar('\n')
#define kg putchar(' ')
using namespace std;
const int maxn=200005,mod=20101009;
int n,a[4],ans[maxn];
char s[maxn][4];
inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
		if(ch=='-')
			f=-1;
		ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
    }
    return x*f;
}
inline void write(int a)
{
    if(a<0)
    {
		putchar('-');
		putchar('1');
    }
    else
    {
		if(a>=10)
			write(a/10);
		putchar(a%10+'0');
    }
}
signed main()
{
	
	n=read();
	a[0]=read();a[1]=read();a[2]=read();
	int flag=0;
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	for(int i=1;i<=n;i++)
	{
		int x=s[i][1]-'A',y=s[i][2]-'A';
		if(a[x]<a[y])
			a[x]++,a[y]--,ans[i]=x;
		else if(a[y]<a[x])
			a[x]--,a[y]++,ans[i]=y;
		else
		{
			if(a[x]==1&&!a[3-x-y])
			{
				if(i==n||(s[i+1][1]!=3-x-y+'A'&&s[i+1][2]!=3-x-y+'A'))
					a[x]--,a[y]++,ans[i]=y;
				else
				{
					int x1=s[i+1][1]-'A',x2=s[i+1][2]-'A';
					if(x1==x||x2==x)
						ans[i]=x,a[x]++,a[y]--;
					else
						ans[i]=y;a[x]--;a[y]++;
				}
			}
			else
				a[x]--;a[y]++;ans[i]=y;
		}
		if(a[x]<0||a[y]<0)
		{
			flag=1;
			break;
		}
	}
	if(flag)
		puts("No");
	else
	{
		puts("Yes");
		for(int i=1;i<=n;i++)
			printf("%c\n",ans[i]+'A');
	}
	return 0;
}