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 <bits/stdc++.h>
using namespace std;
int C[3];
string s[100005];
int ans[100005];
void no(){
    printf("No\n");
    exit(0);
}
int i;
void AB(){ ans[i] = 1; C[0]--; C[1]++; }
void BA(){ ans[i] = 0; C[1]--; C[0]++; }
void AC(){ ans[i] = 2; C[0]--; C[2]++; }
void CA(){ ans[i] = 0; C[2]--; C[0]++; }
void BC(){ ans[i] = 2; C[1]--; C[2]++; }
void CB(){ ans[i] = 1; C[2]--; C[1]++; }
long long x;
int main(){
    int n;
    scanf("%d%d%d%d",&n,&C[0],&C[1],&C[2]);
    x = (long long)C[0]+(long long)C[1]+(long long)C[2];
    for (i = 0; i < n; i++){
        cin >> s[i];
    }
    if (x == 0ll){
        printf("No\n");
        return 0;
    }
    if (x == 1ll){
        for (i = 0; i < n; i++){
            if (s[i] == "AB"){
                if (C[0] == 1) AB();
                else if (C[1] == 1) BA();
                else no();
            }
            if (s[i] == "AC"){
                if (C[0] == 1) AC();
                else if (C[2] == 1) CA();
                else no();
            }
            if (s[i] == "BC"){
                if (C[1] == 1) BC();
                else if (C[2] == 1) CB();
                else no();
            }
        }
        return 0;
    }
    if (s[0] == "AB" && C[0] == 0 && C[1] == 0) no();
    if (s[0] == "BC" && C[1] == 0 && C[2] == 0) no();
    if (s[0] == "AC" && C[0] == 0 && C[2] == 0) no();
    for (i = 0; i < n; i++){
        if (s[i] == "AB"){
            if (C[0] == 0) BA();
            else if (C[1] == 0) AB();
            else if (x == 2ll && C[0] == 1 && C[1] == 1){
                if (i == n-1 || s[i+1] == s[i]){
                    if (C[1] > C[0]) BA();
                    else AB();
                }
                else if (s[i+1] == "BC") AB();
                else BA();
            }
            else if (C[1] > C[0]) BA();
            else AB();
        }
        if (s[i] == "AC"){
            if (C[0] == 0) CA();
            else if (C[2] == 0) AC();
            else if (x == 2ll && C[0] == 1 && C[2] == 1){
                if (i == n-1 || s[i+1] == s[i]){
                    if (C[0] > C[2]) AC();
                    else CA();
                }
                else if (s[i+1] == "BC") AC();
                else CA();
            }
            else if (C[0] > C[2]) AC();
            else CA();
        }
        if (s[i] == "BC"){
            if (C[1] == 0) CB();
            else if (C[2] == 0) BC();
            else if (x == 2ll && C[1] == 1 && C[2] == 1){
                if (i == n-1 || s[i+1] == s[i]){
                    if (C[1] > C[2]) BC();
                    else CB();
                }
                else if (s[i+1] == "AB") CB();
                else BC();
            }
            else if (C[1] > C[2]) BC();
            else CB();
        }
        assert(C[0] >= 0 && C[1] >= 0 && C[2] >= 0);
    }
    printf("Yes\n");
    for (i = 0; i < n; i++){
        printf("%c\n",ans[i]+'A');
    }


}