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; 
const int MAX_N = 2e5 + 5;
int n, a[3];
char s[MAX_N][10];
int ans[MAX_N]; 
int main() {
	scanf("%d", &n); 
	for (int i = 0; i < 3; i++) scanf("%d", a + i); 
    bool 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("Yes"); 
		for (int i = 1; i <= n; i++) printf("%c\n", ans[i] + 'A'); 
	} else puts("No"); 
	return 0;
}