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<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<tuple>
#include<cmath>
#include<iomanip>


#define MOD 1000000007
#define INF 1001001001
#define MIN -1001001001
#define rep(i,k,N) for(int i=k;i<N;i++)
#define MP make_pair
#define MT make_tuple //tie,make_tuple は別物
#define PB push_back
#define PF push_front
#define all(x) (x).begin(),(x).end()


using namespace std;
typedef long long ll;
typedef vector<ll> v;
typedef vector<vector<ll>> vv;

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

//0 0 ?
vector<string> s;
ll N;
vector<char> ans;//
bool ok =false;

bool memo[3][3][3][100001];
void dfs(ll a,ll b,ll c,ll now){
    if(ok)return;
    if(memo[a][b][c][now] == false)return;
    if(now == N){
        cout<<"Yes"<<endl;
        rep(i,0,N){
            cout<<ans[i]<<endl;
        }
        ok = true;
        return;
    }
    if(s[now] == "AB"){
        if(a == 0 && b == 0){
            return;
        }
        else if(a == 0){
            ans[now] = 'A';
            dfs(a+1,b-1,c,now+1);
            ans[now] = '0';
        }
        else if(b == 0){
            ans[now] = 'B';
            dfs(a-1,b+1,c,now+1);
            ans[now] = '0';
        }
        else{
            ans[now] = 'A';
            dfs(a+1,b-1,c,now+1);
            ans[now] = '0';
            ans[now] = 'B';
            dfs(a-1,b+1,c,now+1);
            ans[now] = '0';
        }
    }
    else if(s[now] == "BC"){
        if(b == 0 && c == 0){
            return;
        }
        else if(b == 0){
            ans[now] = 'B';
            dfs(a,b+1,c-1,now+1);
            ans[now] = '0';
        }
        else if(c == 0){
            ans[now] = 'C';
            dfs(a,b-1,c+1,now+1);
            ans[now] = '0';
        }
        else{
            ans[now] = 'B';
            dfs(a,b+1,c-1,now+1);
            ans[now] = '0';
            ans[now] = 'C';
            dfs(a,b-1,c+1,now+1);
            ans[now] = '0';
        }
    }
    else{
        if(a == 0 && c == 0){
            return;
        }
        else if(a == 0){
            ans[now] = 'A';
            dfs(a+1,b,c-1,now+1);
            ans[now] = '0';
        }
        else if(c == 0){
            ans[now] = 'C';
            dfs(a+1,b,c-1,now+1);
            ans[now] = '0';
        }
        else{
            ans[now] = 'A';
            dfs(a+1,b,c-1,now+1);
            ans[now] = '0';
            ans[now] = 'C';
            dfs(a+1,b,c-1,now+1);
            ans[now] = '0';
        }
    }
    memo[a][b][c][now] = false;
}
int main(){
    ll A,B,C;
    cin>>N>>A>>B>>C;
    s.resize(N);
    ans.resize(N);
    rep(i,0,N){
        cin>>s[i];
    }
    if(A+B+C<=2){
        dfs(A,B,C,0);
        if(!ok){
            cout<<"No";
        }
    }
    else if(A == 0 && B == 0 && s[0] == "AB"){
        cout<<"No";
    }
    else if(B == 0 && C == 0 && s[0] == "BC"){
        cout<<"No";
    }
    else if(C == 0 && A == 0 && s[0] == "CA"){
        cout<<"No";
    }
    else{
        rep(i,0,N){
            if(s[i] == "AB"){
                if(A == 0){
                    cout<<"A"<<endl;
                    A++;
                    B--;
                }
                else{
                    cout<<"B"<<endl;
                    B++;
                    A--;
                }
            }
            if(s[i] == "BC"){
                if(C == 0){
                    cout<<"C"<<endl;
                    C++;
                    B--;
                }
                else{
                    cout<<"B"<<endl;
                    B++;
                    C--;
                }
            }
            if(s[i] == "AC"){
                if(A == 0){
                    cout<<"A"<<endl;
                    A++;
                    C--;
                }
                else{
                    cout<<"C"<<endl;
                    C++;
                    A--;
                }
            }
        }
    }
    return 0;
}