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 s[100000];
int n;
int64_t x[3], t;
vector<int> ans;

bool flg = false;
int red;
void solve(int i){
    int nx;
    for(;i<n;++i){
        if(flg){
            if(red == s[i]){
                nx = 3 - s[i] - ans.back();
                flg = false;
            }else{
                int com = 3 - s[i] - red;
                if(com == ans.back()) nx = 3 - s[i] - com;
                else nx = com;
                red = com;
            }
        }else{
            if(x[(s[i]+1)%3] < x[(s[i]+2)%3]) nx = (s[i]+1)%3;
            else nx = (s[i]+2)%3;
            flg = true;
            red = s[i];
        }
        ans.push_back(nx);
    }
    for(int a: ans){
        cout << "ABC"[a] << endl;
    }
}

int main(){
    cin >> n >> x[0] >> x[1] >> x[2];
    t = x[0] + x[1] + x[2];
    for(int i=0;i<n;++i){
        string si;
        cin >> si;
        if(si[0] == 'A' && si[1] == 'B') s[i] = 2;
        if(si[0] == 'B' && si[1] == 'C') s[i] = 0;
        if(si[0] == 'A' && si[1] == 'C') s[i] = 1;
    }
    int o = -1;
    if(t == x[0]) o = 0;
    if(t == x[1]) o = 1;
    if(t == x[2]) o = 2;
    for(int i=0;i<n;++i) cerr << s[i] << " "; cerr << endl;
    
    if(o == -1){
        cout << "Yes" << endl;
        solve(0);
    }else{
        for(int i=0;i<n;++i){
            if(s[i] == o){ cout << "No" << endl; return 0; }
            ans.push_back(3-o-s[i]);
            --x[o];
            ++x[3-o-s[i]];
            if(x[o] > 0){
                cout << "Yes" << endl;
                solve(i+1);
            }
        }
    }

    return 0;
}