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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
const int MR=1e6+10;
typedef long long LL;

int p[3][MR],cnt[3][MR],op[MR];
char ans[MR];

int main() {
	int n,A[3];
	string s;
	cin>>n>>A[0]>>A[1]>>A[2];
	for(int i=1;i<=n;i++){
		cin>>s;
		if(s=="AB") op[i]=0;
		if(s=="BC") op[i]=1;
		if(s=="CA") op[i]=2;		
	}

	for(int i=n;i>=1;i--){
		for(int j=0;j<3;j++){
			if(op[i]==j){
				cnt[j][i]=cnt[j][i+1]+1;
				p[j][i]=i;
			}
			else{
				cnt[j][i]=cnt[j][i+1];
				p[j][i]=p[j][i+1];
			}
		}

	}
	bool flag=true;
	for(int i=1;i<n;i++){
		int x=op[i];
		int y=(x+1)%3;
		int z=(x+2)%3;
		if(A[x]==0 && A[y]==0){
			flag=false;
			break;
		}
		if(A[x]<A[y]){
			ans[i]='A'+x;
			A[x]++;A[y]--;
		}
		else if(A[x]>A[y]){
			ans[i]='A'+y;
			A[x]--;A[y]++;			
		}
		else{
			if(op[i+1]==y){
				ans[i]='A'+y;
				A[x]--;A[y]++;				
			}else if(op[i+1]==z){
				ans[i]='A'+x;
				A[x]++;A[y]--;
			}
			else{
				ans[i]='A'+x;
				A[x]++;A[y]--;				
			}
		}
	}
	int x=op[n];
	int y=(x+1)%3;
	if(A[x]>0){
		ans[n]='A'+y;
	}
	else if(A[y]>0){
		ans[n]='A'+x;
	}else{
		flag=false;
	}
	if(!flag) {
		printf("No\n");
	}
	else{
		printf("Yes\n");
		for(int i=1;i<=n;i++){
			printf("%c\n",ans[i]);
		}		
	}

    return 0;
}