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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, a[3], fs, sc;
bool f;
char s[2][3], ans[100005];
int main(){
	scanf("%d%d%d%d\n%s", &n, a, a+1, a+2, s[1]);
	if(a[s[1][0]-'A'] == 0 && a[s[1][1]-'A'] == 0) return 0 * puts("No");
	rep(i,n){
		if(i < n-1) scanf("%s", s[i&1]);
		fs = s[~i&1][0]-'A';
		sc = s[~i&1][1]-'A';
		if(a[fs] + a[sc] == 0) return 0 * puts("No");
		f = a[fs] > a[sc] || (a[fs] == a[sc] && ('A'+sc == s[i&1][0] || 'A'+sc == s[i&1][1]));
		ans[i] = s[~i&1][f];
		a[f ? sc : fs]++;
		a[f ? fs : sc]--;
	}
	puts("Yes");
	rep(i,n) printf("%c\n", ans[i]);
}